# Use straightforward algebraic methods and solve equations (1.1)

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Use straightforward algebraic methods and solve equations (1.1)
PASS WITH PEARSON Maths Achievement Standard 90147 Use straightforward algebraic methods and solve equations (1.1) Gwenda Hill How to use this CD and Legal Notices Press the ‘Esc’ key to exit at any time Start © Pearson Education New Zealand 2005

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Algebra Main Menu This CD consists of: A Main Menu (this page) returned to by pressing: Three Sections within the main menu accessed by: Exercise groups within each section accessed by: Questions within each exercise group accessed by: Back to Main Menu Back to ‘Section’ Menu Exercise ‘1’ Q1. Achievement Merit Excellence Press to go to desired section. Exit Title Page

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Achievement Menu Press the button to go to the menu page for each exercise group. Exercise A1. Like terms Exercise A2. Simplifying Exercise A3. More simplifying Exercise A4. Expand and simplify Exercise A5. Expand and simplify Exercise A6. Factorise common factors Exercise A7. Factorising quadratics Exercise A8. Algebraic exponent powers Exercise A9. Algebraic exponent factors Exercise A10. Substitution Exercise A11. Finding linear rule Exercise A12. More linear rules Exercise A13. Solve linear equations Exercise A14. More linear equations Exercise A15. Solve factorised quadratics Exercise A16. Achievement review Back to Main Menu

Answer each question then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Are the following pairs like or unlike terms? 4a, 3a Are the following pairs like or unlike terms? 3x, 2x2 Are the following pairs like or unlike terms? 4a2, 3a3 Are the following pairs like or unlike terms? 5p, 3 x p Select the like terms to 6x from: 5x, 7, -2x, x2, 4xy Select the like terms to 2a2 from: 4a, 4a4, -2a2, 5a2, 45a2y Select the like terms to 7 from: 6x, -7, 4, 2a5, 1/4 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Next Back to Achievement Menu

Exercise A1 Question 1 1. Are the following pair like or unlike terms? 4a, 3a Only variable is a, Both have power 1 so Like Next Back to Exercise A1 Menu

Exercise A1 Question 2 2. Are the following pair like or unlike terms? 3x, 2x2 Only variable is x, but One is x and other is x2 so Unlike Previous Next Back to Exercise A1 Menu

Exercise A1 Question 3 3. Are the following pair like or unlike terms? 4a2, 3a3 Only variable is a, One has power 2 and the other 3, so Unlike Previous Next Back to Exercise A1 Menu

Exercise A1 Question 4 4. Are the following pair like or unlike terms? 5p, 3  p Only variable is p and 3  p = 3p Both have power 1, so Like Previous Next Back to Exercise A1 Menu

Exercise A1 Question 5 5. Select the like terms to 6x from: 5x, 7, -2x, x2, 4xy Only variable is x, 5x, -2x, x2, 4xy have x x2 has power 2 not 1, 4xy also has y so 5x and -2x Previous Next Back to Exercise A1 Menu

Exercise A1 Question 6 6. Select the like terms to 2a2 from:
4a, 4a4, -2a2, 5a2, 45a2y Only variable is a with power 2, -2a2, 5a2 and 45a2y have a2 45a2y also has y so -2a2 and 5a2 Previous Next Back to Exercise A1 Menu

Exercise A1 Question 7 7. Select the like terms to 7 from: 6x, -7, 4, 2a5, 1/4 No variable – just a constant -7, 4 and 1/4 are the constants so like terms are: -7, 4 and 1/4 Previous Next Menu Back to Exercise A1 Menu

Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). 4a + 3a 4a2 + 3a3 4a2 – 2a2 + 5a2 5x + 6y + 3x – 2y 3x – 2x 5p – 3p + 7p 6x – 7 – 4x + 2 4x + 2x2 – 3x Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Achievement Menu

Exercise A2 Question 1 1. Simplify the following expression: 4a + 3a
Like terms = 7a Previous Next Back to Exercise A2 Menu

Exercise A2 Question 2 2. Simplify the following expression: 3x – 2x
Like terms = 1x = x Previous Next Back to Exercise 2 Menu

Exercise A2 Question 3 3. Simplify the following expression, where possible: 4a a3 No like terms as different powers So not possible to simplify Previous Next Back to Exercise 2 Menu

Exercise A2 Question 4 4. Simplify the following expression: 5p – 3p + 7p Like terms = 2p + 7p = 9p Previous Next Back to Exercise 2 Menu

Exercise A2 Question 5 5. Simplify the following expression: 4a2 – 2a2 + 5a2 Like terms = 2a2 + 5a2 = 7a2 Previous Next Back to Exercise 2 Menu

Exercise A2 Question 6 6. Simplify the following expression: 6x – 7 – 4x + 2 Like terms together 6x – 4x and = 2x + -5 Previous Next Back to Exercise 2 Menu

Exercise A2 Question 7 7. Simplify the following expression: 5x + 6y + 3x – 2y Like terms together 5x + 3x and 6y – 2y = 8x + 4y Previous Next Back to Exercise 2 Menu

Exercise A2 Question 8 8. Simplify the following expression: 4x + 2x2 – 3x Like terms together 4x – 3x = x + 2x2 Previous Next Menu Back to Exercise 2 Menu

Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). 4a  3a 4a2  3a3 4a2  -2a3  a2 5x  6y  3x  2y 3x  -2x 5p  3p2  p 6x  -7x  2 4x  2x2  3xy Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Achievement Menu

Exercise A3 Question 1 1. Simplify the following expression: 4a  3a
4  3 = 12, a  a = a2 so 12a2 Next Back to Exercise A3 Menu

Exercise A3 Question 2 2. Simplify the following expression: 3x  -2x
3  -2 = -6, x  x = x2 so -6x2 Previous Next Back to Exercise A3 Menu

Exercise A3 Question 3 3. Simplify the following expression: 4a2  3a3
4  3 = 12, a2  a3 = a5 so 12a5 Previous Next Back to Exercise A3 Menu

Exercise A3 Question 4 4. Simplify the following expression: 5p  3p2  p 5  3  1 = 15, p  p2  p = p1+2+1 = p4 so 15p4 Previous Next Back to Exercise A3 Menu

Exercise A3 Question 5 5. Simplify the following expression: 4a2  -2a3  a2 4  -2  1 = -8, a2  a3  a2 = a7 so -8a7 Previous Next Back to Exercise A3 Menu

Exercise A3 Question 6 6. Simplify the following expression: 6x  -7x  2 6  -7  2 = -84, x  x = x2 so -84x2 Previous Next Back to Exercise A3 Menu

Exercise A3 Question 7 7. Simplify the following expression: 5x  6y  3x  2y 5  6  3  2 = 180, x  x = x2, y  y = y2 so 180x2 y2 Previous Next Back to Exercise A3 Menu

Exercise A3 Question 8 8. Simplify the following expression: 4x  2x2  3xy 4  2  3 = 24, x  x2  x = x4 so 24x4 y Previous Next Menu Back to Exercise A3 Menu

Exercise A4 Menu Expand and simplify
Expand and simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). 4 (a + 3) a (a + 5) 4a (2a + 3) 5x (6y – 3x) 2x2 (3xy2 + 2x2y) 3 (x – 2) p (p – 3) 6x ( -7x + 2) -4x ( 2x – 3y) 3x2y (xy2 – 2x3y) Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10. Next Back to Achievement Menu

Exercise A4 Question 1 1. Expand and simplify: 4 (a + 3)
4  a = 4a, 4  3 = 12 so 4a + 12 Next Back to Exercise A4 Menu

Exercise A4 Question 2 2. Expand and simplify: 3 (x – 2)
3  x = 3x, 3  2 = 6 so 3x – 6 Previous Next Back to Exercise A4 Menu

Exercise A4 Question 3 3. Expand and simplify: a (a + 5)
a  a = a2, a  5 = 5a so a2 + 5a Previous Next Back to Exercise A4 Menu

Exercise A4 Question 4 4. Expand and simplify: p (p – 3)
p  p = p2, p  3 = 3p so p2 – 3p Previous Next Back to Exercise A4 Menu

Exercise A4 Question 5 5. Expand and simplify: 4a (2a + 3)
4a  2a = 8a2, 4a  3 = 12a so 8a2 + 12a Previous Next Back to Exercise A4 Menu

Exercise A4 Question 6 6. Expand and simplify: 6x (-7x + 2)
6x  -7x = -42x2, 6x  2 = 12x so -42x2 + 12x or 12x – 42x2 Previous Next Back to Exercise A4 Menu

Exercise A4 Question 7 7. Expand and simplify: 5x (6y – 3x)
5x  6y = 30xy, 5x  3x = 15x2 so 30xy – 15x2 Previous Next Back to Exercise A4 Menu

Exercise A4 Question 8 8. Expand and simplify: -4x (2x – 3y)
-4x  2x = -8x2, -4x  3y = -12xy so -8x2 – -12xy therefore -8x2 + 12xy or 12xy – 8x2 Previous Next Back to Exercise A4 Menu

Exercise A4 Question 9 9. Expand and simplify: 2x2 (3xy2 + 2x2y)
2x2  3xy2 = 6x3y2, 2x2  2x2y = 4x4y so 6x3y2 + 4x4y Previous Next Back to Exercise A4 Menu

Exercise A4 Question 10 10. Expand and simplify: 3x2y (xy2 – 2x3y)
3x2y  xy2 = 3x3y3, 3x2y  2x3y = 6x5y2 so 3x3y3 + 6x5y2 Previous Next Menu Back to Exercise A4 Menu

Exercise A5 Menu Expand and simplify
Expand and simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). (a + 3) (a + 4) (a – 3) (a + 5) (a + 4) (2a + 3) (5x + y) (x – y) (a – b) (a + b) (a + b)2 (x + 5) (x + 2) (p – 6) (p – 3) (x – 3) (-7x + 2) (x + 4) (2x – 3) (x – 3)2 (2x – 5) (2x + 5) Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10. Q11. Q12. Next Back to Achievement Menu

Exercise A5 Question 1 1. Expand and simplify: (a + 3) (a + 4)
First Outside Inside Last aa=a2 a4=4a 3a=3a 34=12 so a2 + 4a + 3a + 12 = a2 + 7a + 12 Next Back to Exercise A5 Menu

Exercise A5 Question 2 2. Expand and simplify: (x + 5) (x + 2)
First Outside Inside Last xx=x 2 x2=2x 5x=5x 52=10 so x2 + 2x + 5x + 10 = x2 + 7x + 10 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 3 3. Expand and simplify: (a – 3) (a + 5)
First Outside Inside Last aa=a2 a5=5a -3a=-3a -35=-15 so a2 + 5a – 3a – 15 = a2 + 2a – 15 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 4 4. Expand and simplify: (p – 6) (p – 3)
First Outside Inside Last pp=p2 p-3=-3p -6p=-6p -6–3=18 so p2 – 3p – 6p + 18 = p2 – 9p + 18 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 5 5. Expand and simplify: (a + 4) (2a + 3)
First Outside Inside Last a2a=2a2 a3=3a 42a=8a 43=12 so 2a2 + 3a + 8a + 12 = 2a2 + 11a + 12 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 6 6. Expand and simplify: (x – 3) (-7x + 2)
First Outside Inside Last x-7x=-7x2 x2=2x -3x –7x=21x -32=-6 so -7x2 + 2x + 21x – 6 = -7x2 + 23x – 6 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 7 7. Expand and simplify: (5x + y) (x – y)
First Outside Inside Last 5xx =5x2 5x-y=-5xy yx=xy y-y=-y2 so 5x2 – 5xy + xy + -y2 = 5x2 – 4xy – y2 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 8 8. Expand and simplify: (x + 4) (2x – 3)
First Outside Inside Last x2x=2x2 x-3=-3x 42x=8x 4-3=-12 so 2x2 – 3x + 8x – 12 = 2x2 + 5x – 12 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 9 9. Expand and simplify: (a – b) (a + b)
First Outside Inside Last aa=a2 ab=ab -ba=-ab -bb=-b2 so a2 + ab – ab – b2 = a2 – b2 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 10 10. Expand and simplify: (x – 3)2
= (x–3) (x–3) First Outside Inside Last xx=x2 x–3=-3x -3x=-3x -3-3=9 so x2 – 3x – 3x + 9 = x2 – 6x + 9 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 11 11. Expand and simplify: (a + b)2
= (a + b) (a + b) First Outside Inside Last aa=a2 ab=ab ba=ab bb=b2 so a2 + ab + ab + b2 = a2 + 2ab + b2 Previous Next Back to Exercise A5 Menu

Exercise A5 Question 12 12. Expand and simplify: (2x – 5)(2x + 5)
First Outside Inside Last 2x2x=4x2 2x5=10x -52x=-10x -55=-25 so 4x2 + 10x – 10x – 25 = 4x2 – 25 Previous Next Menu Back to Exercise A5 Menu

Exercise A6 Menu Factorising common factors
Fully factorise the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). 3a – 15 a2 + 4a 2xy + 2xt 12x + 3p 15a2b3 + 25ab2 4x + 24 p – 6pq 3q + 15q2 9x2y + 30xy2 30x2y – 21x2y3 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10. Next Back to Achievement Menu

Exercise A6 Question 1 1. Factorise fully: 3a – 15 = 3  a – 3  5
Next Back to Exercise A6 Menu

Exercise A6 Question 2 2. Factorise fully: 4x + 2 = 4  x + 4  6
Previous Next Back to Exercise A6 Menu

Exercise A6 Question 3 3. Factorise fully: a2 + 4a = a  a + a  4
Previous Next Back to Exercise A6 Menu

Exercise A6 Question 4 4. Factorise fully: p – 6pq = p  1 – p  6q
Previous Next Back to Exercise A6 Menu

Exercise A6 Question 5 5. Factorise fully: 2xy + 2xt
= 2x (y + t) Previous Next Back to Exercise A6 Menu

Exercise A6 Question 6 6. Factorise fully: 3q + 15q2
= 3  q + 3  5  q  q = 3q (1 + 5q) Note: Must have the 1 Previous Next Back to Exercise A6 Menu

Exercise A6 Question 7 7. Factorise fully: 12x + 3p = 3  4x + 3  p
Previous Next Back to Exercise A6 Menu

Exercise A6 Question 8 8. Factorise fully: 9x2y + 30xy2
= 3  3  x  x  y + 3  10  x  y  y Common factors are 3, x, y = 3xy(3x + 10y) Previous Next Back to Exercise A6 Menu

Exercise A6 Question 9 9. Factorise fully: 15a2b3 + 25ab2
= 3  5  a  a  b  b  b + 5  5  a  b  b Common factors are 5, a, b and b = 5ab2 (3ab + 5) Previous Next Back to Exercise A6 Menu

Exercise A6 Question 10 10. Factorise fully: 30x2y – 21x2y3
= 3  10  x  x  y – 3  7  x  x  y  y  y Common factors are 3, x2 and y = 3x2y (10 – 7y2) Previous Next Menu Back to Exercise A6 Menu

Fully factorise the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). a2 + 4a + 3 a2 + 4a – 5 a2 + 6a – 16 x2 + x – 20 a2 – 9 x2 + 4x + 4 x2 – 4x – 5 p2 – 6p + 8 k2 – 6k – 7 a2 + 6a + 9 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10. Next Back to Achievement Menu

Exercise A7 Question 1 1. Factorise fully: a2 + 4a + 3
No common factors Pairs that multiply to give 3 are 1, 3 and -1, -3 1 + 3 = 4 so (a + 1)(a + 3) or (a + 3)(a + 1) Next Back to Exercise A7 Menu

Exercise A7 Question 2 2. Factorise fully: x2 + 4x + 4
No common factors Pairs that multiply to give 4 are: 1, 4; 2, 2; -1, -4 and -2, -2 2 + 2 = 4 so (x + 2)(x + 2) = (x + 2)2 Previous Next Back to Exercise A7 Menu

Exercise A7 Question 3 3. Factorise fully: a2 + 4a – 5
No common factors Pairs that multiply to give -5 are: 1, -5 and 5, -1 = 4 so (a + 5)(a – 1) or (a – 1)(a + 5) Previous Next Back to Exercise A7 Menu

Exercise A7 Question 4 4. Factorise fully: x2 – 4x – 5
No common factors Pairs that multiply to give -5 are: 1, -5 and 5, -1 = -4 so (x + 1)(x – 5) or (x – 5)(x + 1) Previous Next Back to Exercise A7 Menu

Exercise A7 Question 5 5. Factorise fully: a2 + 6a – 16
No common factors Pairs that multiply to give -16 are: 1, -16; 2, -8; 4, -4; 8, -2 and 16, -1 = 6 so (a + 8)(a – 2) or (a – 2)(a + 8) Previous Next Back to Exercise A7 Menu

Exercise A7 Question 6 6. Factorise fully: p2 – 6p + 8
No common factors Pairs that multiply to give 8 are: 1, 8; 2, 4; -1, -8 and -2, -4 = -6 so (p – 2)(p – 4) or (p – 4)(p – 2) Previous Next Back to Exercise A7 Menu

Exercise A7 Question 7 7. Factorise fully: x2 + x – 20
No common factors Pairs that multiply to give -20 are: 1, -20; 2, -10; 4, -5; 5, -4; 10, -2 and 20, -1 x means 1x and = 1 so (x + 5)(x – 4) or (x – 4)(x + 5) Previous Next Back to Exercise A7 Menu

Exercise A7 Question 8 8. Factorise fully: k2 – 6k – 7
No common factors Pairs that multiply to give -7 are: 1, -7 and 7, -1 = -6 so (k + 1)(k – 7) or (k – 7)(k + 1) Previous Next Back to Exercise A7 Menu

Exercise A7 Question 9 9. Factorise fully: a2 – 9
No common factors Can be written as: a2 + 0p – 9 Pairs that multiply to give -9 are: 1, -9; 3, -3 and 9, -1 = 0 so (a + 3)(a – 3) or (a – 3)(a + 3) Previous Next Back to Exercise A7 Menu

Exercise A7 Question 10 10. Factorise fully: a2 + 6a + 9
No common factors Pairs that multiply to give 9 are: 1, 9; 3, 3; -1, -9 and 3, -3 3 + 3 = 6 so (a + 3)(a + 3) = (a + 3)2 Previous Next Menu Back to Exercise A7 Menu

Exercise A8 Menu Algebraic exponent powers
Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). (3x)2 (4y)3 Q1. Q2. (5a2)2 (2x2y)4 Q3. Q4. Next Back to Achievement Menu

Exercise A8 Question 1 1. Simplify: (3x)2 = 3x  3x = 9x2 Next

Exercise A8 Question 2 2. Simplify: (4y)3 = 4y  4y  4y = 43y3 = 64y3
Previous Next Back to Exercise A8 Menu

Exercise A8 Question 3 3. Simplify: (5a2)2 = 5a2  5a2 = 52(a2)2
Previous Next Back to Exercise A8 Menu

Exercise A8 Question 4 4. Simplify: (2x2y)4 = 24  (x2)4  (y)4

Exercise A9 Menu Algebraic exponent factors
Simplify the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Achievement Menu

Exercise A9 Question 1 1. Simplify:
Common factors on top and bottom are: 4 and x 4xy = 4x  y 8x = 4x  2 Next Back to Exercise A9 Menu

Exercise A9 Question 2 2. Simplify:
Common factors on top and bottom are: 3 and x 12x = 3x  4 15xy = 3x  5y Previous Next Back to Exercise A9 Menu

Exercise A9 Question 3 3. Simplify:
Common factors on top and bottom are: 3 and x 3x2 = 3x  x 6xy = 3x  2y Previous Next Back to Exercise A9 Menu

Exercise A9 Question 4 4. Simplify:
Common factors on top and bottom are: 3, x2 and y2 12x3y2 = 3x2 y2  4x 15x2y3 = 3x2 y2  5y Previous Next Back to Exercise A9 Menu

Exercise A9 Question 5 5. Simplify:
Common factors on top and bottom are: 12 and a (Note: It does not matter if you only notice one or a smaller factor – this just means you will take more steps as you find other common factors) 36a2 = 12a  3a 60ab2 = 12a  5b2 Previous Next Back to Exercise A9 Menu

Exercise A9 Question 6 6. Simplify:
Common factors on top and bottom are: 2 and x2 2x2 = 2x2  1, 4x2y = 2x2  2y Previous Next Back to Exercise A9 Menu

Exercise A9 Question 7 7. Simplify:
Common factors on top and bottom are: 12, p and q 36p2q4 = 12pq  3pq3, 60pq = 12pq  5 Previous Next Back to Exercise A9 Menu

Exercise A9 Question 8 8. Simplify: Top is 25x2
Common factors on top and bottom are: 5 and x 25x2 = 5x  5x, 60xy3 = 5x  12y3 Previous Next Menu Back to Exercise A9 Menu

Substitute within the following expressions then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Q1. If A = 0.5(a + b)h, find A when a = 3, b = 7 and h = 6 If k = 0.6b2, find k when b = 4 S = What is S when a = 3, c = 5 and d = 2? The volume of a sphere is found by where r is radius. Find the volume of a sphere with radius 5. A new symbol  has the meaning that x  y = 3x2 – y. Find the value of -2  4. To convert Fahrenheit temperature, f, to Celsius, C, use the formula Convert 360o Fahrenheit to oCelsius. Q2. Q3. Q4. Q5. Q6. Next Back to Achievement Menu

Exercise A10 Question 1 1. If A = 0.5(a + b)h, find A when a = 3, b = 7 and h = 6 A = 0.5  (3 + 7)  6 = 30 Next Back to Exercise A10 Menu

Exercise A10 Question 2 2. If k = 0.6b2, find k when b = 4
= 9.6 Previous Next Back to Exercise A10 Menu

Exercise A10 Question 3 3. S = What is S when a = 3, c = 5 and d = 2? = (15  3)  (5 – 2) = 15 Previous Next Back to Exercise A10 Menu

Exercise A10 Question 4 4. The volume of a sphere is found by where r is radius. Find the volume of a sphere with radius 5. V = = 4  3    53 = = 524 to 3 sig fig Note: There was only 1 sig fig in the radius, so it might be acceptable to write 500 to 1 sig fig. However, you must show working steps and it is a good idea to show the unrounded answer. Previous Next Back to Exercise A10 Menu

Exercise A10 Question 5 5. A new symbol  has the meaning that x  y = 3x2 – y. Find the value of -2  4 -2 takes the place of x and 4 takes the place of y -2  4 = 3(-2)2 – 4 Note: Use brackets to square a negative = 8 Previous Next Back to Exercise A10 Menu

Exercise A10 Question 6 = 182.222222 so 182.2 Celsius to 1 d.p.
6. To convert Fahrenheit temperature, f, to Celsius, C, use the formula Convert 360o Fahrenheit to o Celsius. = so Celsius to 1 d.p. Note: Other rounding acceptable. Previous Next Menu Back to Exercise A10 Menu

Exercise A11 Menu Finding linear rule
Find the rule for finding the nth term for the following sequences then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). 8, 12, 16, 20, … 10, 17, 24, 31, … 24, , 34.5, … 35, 32, 29, 26, … 22, 24, 26, 28, … 15, 15.5, 16, 16.5, … 104, 102, 100, 98, … 12.5, 12.1, 11.7, 11.3, ... Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Achievement Menu

Exercise A11 Question 1 1. Find the rule for finding the nth term for the sequence: 8, 12, 16, 20, … Difference between terms is 4 Rule is from the 4 x table 4, 8, 12, 16, … Need 4 more to give the sequence Therefore rule is 4n + 4 Next Back to Exercise A11 Menu

Exercise A11 Question 2 2. Find the rule for finding the nth term for the sequence: 22, 24, 26, 28, … Difference between terms is 2 Rule is from the 2 x table 2, 4, 6, 8, … Need 20 more to give the sequence Therefore rule is 2n + 20 Previous Next Back to Exercise A11 Menu

Exercise A11 Question 3 3. Find the rule for finding the nth term for the sequence: 10, 17, 24, 31, … Difference between terms is 7 Rule is from the 7 x table 7, 14, 21, 28, … Need 3 more to give the sequence Therefore rule is 7n + 3 Previous Next Back to Exercise A11 Menu

Exercise A11 Question 4 4. Find the rule for finding the nth term for the sequence: 15, 15.5, 16, 16.5, … Difference between terms is 0.5 Multiplying 1, 2, 3, 4, … by 0.5 gives: 0.5, 1, 1.5, 2, … Need 14.5 more to give the sequence Therefore rule is 0.5n Previous Next Back to Exercise A11 Menu

Exercise A11 Question 5 5. Find the rule for finding the nth term for the sequence: 24, , 34.5, … Difference between terms is 3.5 Multiplying 1, 2, 3, 4, … by 3.5 gives: 3.5, 7, 10.5, 14, … Need 20.5 more to give the sequence Therefore rule is 3.5n Previous Next Back to Exercise A11 Menu

Exercise A11 Question 6 6. Find the rule for finding the nth term for the sequence: 104, 102, 100, 98, … Difference between terms is -2 Multiplying 1, 2, 3, 4, … by -2 gives: -2, -4, -6, -8, … Need 106 more to give the sequence Therefore rule is -2n or 106 – 2n Previous Next Back to Exercise A11 Menu

Exercise A11 Question 7 7. Find the rule for finding the nth term for the sequence: 35, 32, 29, 26, … Difference between terms is -3 Multiplying 1, 2, 3, 4, … by -3 gives: -3, -6, -9, -12, … Need 38 more to give the sequence Therefore rule is -3n or 38 – 3n Previous Next Back to Exercise A11 Menu

Exercise A11 Question 8 8. Find the rule for finding the nth term for the sequence: 12.5, 12.1, 11.7, 11.3, ... Difference between terms is –0.4 Multiplying 1, 2, 3, 4, … by –0.4 gives: –0.4, –0.8, –1.2, –1.6, … Need 12.9 more to give the sequence Therefore rule is –0.4n or – 0.4n Previous Next Menu Back to Exercise A11 Menu

Exercise A12 Menu Linear rules diagram or table
Find the rule for finding the nth term for the following sequences then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. A matches design. Find a rule and work out how many matches would be needed. A rectangular shape. Find a rule to find the perimeter of the 10th rectangle that would be drawn in this sequence. Q1. Q2. The telephone account. Complete the table to show the monthly costs. Write a rule to show how much is paid for n calls. Q3. A rectangular courtyard is being tiled. Find a rule to find the number of tiles after n edges have been added. Q4. The savouries table. Find an expression for the number of savouries allowed for n guests. The amount of sewing pattern material. Write a general rule to give the amount of material for width (w) Q5. Q6. The number of doors needed. If a block of flats is to have 17 storeys, how many doors are needed? Q7. Next Back to Achievement Menu

Exercise A12 Question 1 Sequence is 7, 12, 17, … Difference is 5
1. The following design is being created with matches. Find a rule and work out how many matches would be needed to make a design with 20 of the basic shapes. Sequence is 7, 12, 17, … Difference is 5 5 x table would give: 5, 10, 15, … Need 2 more each time so 5n + 2 20 basic shapes, n = 20, Need 5  = 102 matches Add figure Ch 01.08 Answer Next Back to Exercise A12 Menu

Exercise A12 Question 2 Sequence is 18, 26, 34, Difference is 8
2. A rectangular shape is being increased by adding 1 metre to each side at the corner as shown. It starts as size 5 m by 4 m so has perimeter of 18 m. Find a rule to find the perimeter of the 10th rectangle that would be drawn in this sequence. Sequence is 18, 26, 34, Difference is 8 8 x table is 8, 16, 24, … Need 10 more each time. Rule is 8n + 10 10th rectangle would have n = 10 Therefore 8  = 90 Answer Add figure Ch 01.09 Previous Next Back to Exercise A12 Menu

Exercise A12 Question 3 Complete table with 28.25, 31, 33.75
Calls 1 2 3 4 5 Cost 22.75 25.50 3. Steven pays \$20 a month for the telephone account. Phone calls to his girlfriend in another town cost \$2.75 each. Complete the table to show the monthly cost for different numbers of phone calls. Write a rule to show how much he would pay for n calls. Complete table with 28.25, 31, 33.75 Difference = 2.75 This gives 2.75, 5.5, 8.25, … Needs 20 added Rule is 2.75n + 20 Answer Previous Next Back to Exercise A12 Menu

Exercise A12 Question 4 Number of tiles is 21, 27, 33, …
ANSWER 4. A rectangular courtyard is being tiled. Fifteen tiles form the rectangular centre as shown and the tiles are being placed on each side. Find a rule to find the number of tiles after n edges have been added. Number of tiles is 21, 27, 33, … Difference is 6 6 x table gives: 6, 12, 18, … Need to add 15 to each Therefore rule is 6n + 15 Add figure Ch 01.10 Answer Previous Next Back to Exercise A12 Menu

Exercise A12 Question 5 Guests go up by 10
5. The table shows the number of savouries allowed by a catering company for different numbers of guests. Find an expression for the number of savouries allowed for n guests. Guests 10 20 30 40 Savouries 25 45 65 85 Guests go up by 10 Savouries go up by 20, so 2 savouries per person. But for 10, 20, 30, 40, this would give: 20, 40, 60, 80, so need to add 5 Therefore rule is 2n + 5 Answer Previous Next Back to Exercise A12 Menu

Exercise A12 Question 6 Waist goes up by 5.
Waist (cm) 70 75 80 85 90 Material (m) 2.7 2.9 3.1 3.3 3.5 6. The amount of material needed for a sewing pattern is given for different sizes as shown in the table. Write a general rule to give the amount of material for the width (w). Waist goes up by 5. Material goes up by 0.2, so 0.04 metres per cm. But for 70, 75, 80, 85, this would give: 2.8, 3, 3.2, 3.4, so need to take 0.1 Therefore rule is 0.04w – 0.1 Answer Previous Next Back to Exercise A12 Menu

Exercise A12 Question 7 Number of doors is 27, 51, 75, 99, …
7. The number of doors needed to be supplied for a block of flats depends on the number of storeys high the flats are. For one storey 27 doors are needed. If the building is two-storied there are 51 doors needed; three-storied, 75 doors; four-storied, 99 doors and so on. If a block of flats is to have 17 storeys, how many doors are needed? Number of doors is 27, 51, 75, 99, … Difference is 24 24 x table gives 24, 48, 72, 96, … Need to add 3 to each Therefore rule is 24n + 3 For 17 storeys, n = 17 so 24  = 411 doors. Answer Add figure Ch 01.10 Previous Next Menu Back to Exercise A12 Menu

Exercise A13 Menu Solve linear equations
Solve the following equations showing working, then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Q1. 5x + 7 = 42 2(x + 7) = 28 26 = 3x – 2 Q2. 3x – 8 = 82 5x + 19 = 32.4 4 – x = 16 Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Achievement Menu

Exercise A13 Question 1 1. Solve the following equation showing working: 5x + 7 = 42 5x + 7 = 42 – 7 – 7 5x = 35 5 5 x = 7 Insert Figure Ch 01.16A at top Figure Ch 01.18B below Next Back to Exercise A13 Menu

Exercise A13 Question 2 2. Solve the following equation showing working: 3x – 8 = 82 3x – 8 = 82 3x = 90  3 x = 30 Figure Ch 01.17A at top Figure Ch 01.17B below Previous Next Back to Exercise A13 Menu

Exercise A13 Question 3 3. Solve the following equation showing working: 2(x + 7) = 28 Note: Can divide by 2 or expand brackets as first step. This example shows expanding brackets. 2x = 28 – – 14 2x = 14   2 x = 7 Previous Next Back to Exercise A13 Menu

Exercise A13 Question 4 4. Solve the following equation showing working: 5x + 19 = 32.4 5x + 19 = 32.4 – – 19 5x = 13.4   5 x = 2.68 Previous Next Back to Exercise A13 Menu

Exercise A13 Question 5 x = 9.333333 so 9.33 (to 2 d.p.)
5. Solve the following equation showing working: 26 = 3x – 2 26 = 3x – 2 28 = 3x 3x = 28 Usual to have x term on left side so swap.   3 Can swap sides at any time since sides equal x = so 9.33 (to 2 d.p.) Previous Next Back to Exercise A13 Menu

Exercise A13 Question 6 6. Solve the following equation showing working: 4 – x = 16 4 – x = 16 – – 4 -x = 12  -1 x = -12 Previous Next Back to Exercise A13 Menu

Exercise A13 Question 7 7. Solve the following equation showing working: – – 3  8  8 x = 192 Previous Next Back to Exercise A13 Menu

Exercise A13 Question 8 8. Solve the following equation showing working: x 8 x 8 x = 24 Previous Next Menu Back to Exercise A13 Menu

Exercise A14 Menu Solve more linear equations
Solve the following equations showing working, then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. 3.6x – 8 = 2.4x 5x + 19 = 30x + 104 14 – 2x = 16x 5x + 7x = 42 2.5(x + 7) = 27.5 5.6x = 1.8x + 7 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Achievement Menu

Exercise A14 Question 1 1. Solve the following equation showing working: 5x + 7x = 42 12x = 42 12  12 x = 3.5 Next Back to Exercise A14 Menu

Exercise A14 Question 2 2. Solve the following equation showing working: 3.6x – 8 = 2.4x 3.6x – 8 = 2.4x – 3.6x – 3.6x -8 = -1.2x -1.2x = Can swap sides at any time   -1.2 x = (to 2 d.p.) Previous Next Back to Exercise A14 Menu

Exercise A14 Question 3 2.5(x + 7) = 27.5
3. Solve the following equation showing working: 2.5(x + 7) = 27.5 Can be done by expanding bracket or dividing by 2.5 This example is done by dividing by 2.5. 2.5(x + 7) = 27.5   2.5 x = 11 – 7 – 7 x = 4 Previous Next Back to Exercise A14 Menu

Could take any term from both sides as a correct first step.
Exercise A14 Question 4 4. Solve the following equation showing working: 5x + 19 = 30x + 104 – 19 – 19 Could take any term from both sides as a correct first step. 5x = 30x + 85 – 30x – 30x -25x = 85   -25 x = -3.4 Previous Next Back to Exercise A14 Menu

Exercise A14 Question 5 5. Solve the following equation showing working: 5.6x = 1.8x + 7 5.6x = 1.8x + 7 – 1.8x – 1.8x 3.8x = 7   3.8 x = (to 2 d.p.) Previous Next Back to Exercise A14 Menu

Exercise A14 Question 6 6. Solve the following equation showing working: 14 – 2x = 16x 14 – 2x = 16x + 2x + 2x 14 = 18x 18x = 14  18  18 x = (to 2 d.p.) Previous Next Back to Exercise A14 Menu

Exercise A14 Question 7 7. Solve the following equation showing working: Multiply all terms by 5 3x – 15 = 10x Note that each term has been  5 – 3x – 3x -15 = 7x 7x = Can swap sides at any time  7  7 x = (to 2 d.p.) Previous Next Back to Exercise A14 Menu

Exercise A14 Question 8 x + 18 = -30x Note that each term has been  6
8. Solve the following equation showing working: Multiply all terms by 6 x = -30x Note that each term has been  6 – x – x 18 = -31x -31x = Can swap sides at any time   -31 x = (to 2 d.p.) Previous Next Menu Back to Exercise A14 Menu

Solve the following equations showing working, then press the button to see the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. (x – 3)(x – 2) = 0 (x + 1)(x + 4) = 0 (3x + 2)(x – 2) = 0 x(2x + 7) = 0 (x – 5)(x + 3) = 0 (2x – 5)(x + 4) = 0 (3x + 2)(2x – 5) = 0 5x (3x – 7) = 0 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Achievement Menu

Exercise A15 Question 1 1. Solve the following equation showing working: (x – 3)(x – 2) = 0 x – 3 = or x – 2 = 0 x = x = 2 Next Back to Exercise A15 Menu

Exercise A15 Question 2 2. Solve the following equation showing working: (x – 5)(x + 3) = 0 x – 5 = or x = 0 – – 3 x = x = -3 Previous Next Back to Exercise A15 Menu

Exercise A15 Question 3 3. Solve the following equation showing working: (x + 1)(x + 4) = 0 x + 1 = or x + 4 = 0 – – 1 – – 4 x = x = -4 Previous Next Back to Exercise A15 Menu

Exercise A15 Question 4 4. Solve the following equation showing working: (2x – 5)(x + 4) = 0 2x – 5 = or x + 4 = 0 – 4 – 4 2x = x = -4  2  2 x = 2.5 Previous Next Back to Exercise A15 Menu

Exercise A15 Question 5 5. Solve the following equation showing working: (3x + 2)(x – 2) = 0 3x = or x – 2 = 0 – – 3x = x = 2   3 x = Previous Next Back to Exercise A15 Menu

Exercise A15 Question 6 6. Solve the following equation showing working: (3x + 2)(2x – 5) = 0 3x = or x – 5 = 0 – 2 – 3x = -2 2x = 5     2 x = (2 d.p.) or x = 2.5 Previous Next Back to Exercise A15 Menu

Exercise A15 Question 7 7. Solve the following equation showing working: x(2x + 7) = 0 x = or 2x + 7 = 0 – – 7 2x = -7   2 x = -3.5 Previous Next Back to Exercise A15 Menu

Exercise A15 Question 8 8. Solve the following equation showing working: 5x (3x – 7) = 0 5x = or 3x – 7 = 0   x = x = 7   3 x = (2 dp) Previous Next Menu Back to Exercise A15 Menu

Click on the buttons to go to the review questions, solve each one, then press the answer button. Or scroll to each question using the next arrow (below). Click to move from step to step in the solution. Q1. Expand the equations Find the rule Solve the equation Q2. Factorise the equations Find the surface area Solve the equations Q3. Q4. << figure Ch01.11>> Q5. Q6. Q7. Next Back to Achievement Menu

Exercise A16 Question 1 1. Expand and simplify:
(a) 3xy (2x – 7) (b) (2x + 3)(x – 6) (a) 3xy  2x – 3xy  7 = 6x2y – 21xy Answer a (b) First Outside Inside Last 2x  x – 2x   x – 3  6 = 2x2 – 12x + 3x – 18 = 2x2 – 9x – 18 Answer b Next Back to Exercise A16 Menu

Exercise A16 Question 2 2. Factorise fully:
(a) 5x2y – 20x3y4 (b) x2 – 8x + 15 (a) = 5  x  x  y – 5  4  x  x  x  y  y  y  y = 5x2y(1 – 4xy3) Answer a (b) Pairs that multiply to give 15 are: 1, 15; 3, 5; -1, and -3, -5 -3 and -5 add to give -8 x2 – 8x + 15 = (x – 3)(x – 5) or (x – 5)(x – 3) Answer b Previous Next Back to Exercise A16 Menu

Exercise A16 Question 3 3. Simplify: (a) (b)
(a) = 53  (x2 )3  22  x2  y2 = 125  4  x6  x2  y2 = 500 x8y2 Answer a (b) Top = 9  p  p  p  p  q  q Bottom = 9  p Common factor is 9p Answer b Previous Next Back to Exercise A16 Menu

Exercise A16 Question 4 S. A. = 4    8  8 = 804.2477193
4. S.A. = gives the surface area of a sphere for given radius r. What is the surface area of a sphere with radius 8 cm? S. A. = 4    8  8 = so 804 (3 sig fig) Answer Previous Next Back to Exercise A16 Menu

Exercise A16 Question 5 5. Hedge plants are to be
placed around the edge of rectangular fields. The diagram shows the spacing for different 2 m wide rectangular fields. Find a rule to give the number of plants needed for a 2 m rectangular field that is n metres long. For n values 0.5, 1, 1.5,… the number of plants is 10, 12, 14 so that the values 1, 2, 3, … would give 12, 16, 20 With difference 4, the 4 x table gives 4, 8, 12, … Therefore need to add 8. The rule for length n is 4n + 8 Answer Previous Next Back to Exercise A16 Menu

Exercise A16 Question 6 6. Solve: (a) 5x – 7 = 32 (b) 6x – 19 = 3.6x + 21 (b) 6x – 19 = 3.6x + 21 6x = 3.6x + 40 – 3.6x – 3.6x 2.4x = 40  2.4 x = (2 d.p.) (a) 5x – 7 = 32 5x = 39   5 x = 7.8 Answer a Answer b Previous Next Back to Exercise A16 Menu

Exercise A16 Question 7 7. Solve: (3x + 5)(6 – x) = 0
3x + 5 = or – x = 0 – – 5 x = 6 3x = -5   3 x = (2 d.p.) Back to Main Menu Previous Merit Menu Back to Exercise A16 Menu

Press the button to go to the menu page for each exercise group.
Merit Menu Press the button to go to the menu page for each exercise group. Exercise M1. Rational expressions Exercise M2. Simplifying Exercise M3. Quadratic patterning Exercise M4. Change the subject Exercise M5. Solving inequations Exercise M6. Equations and inequations Exercise M7. Simple quadratic equations Exercise M8. Harder quadratic equations Exercise M9. Elimination in equations Exercise M10. Substitution in equations Exercise M11. Context equation solving Back to Main Menu

Answer each question as a single expression. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Next Back to Merit Menu

Exercise M1 Question 1 1. Write as a single expression:
3 and 5 have common multiple of 15 Rewrite x/3 as 5x/15 by multiplying top and bottom by 5 Rewrite x/5 as 3x/15 by multiplying top and bottom by 3 Write common bottom line and combine the top lines. Next Back to Exercise M1 Menu

Exercise M1 Question 2 2. Write as a single expression:
4 and 7 have common multiple of 28 Rewrite x/4 as 7x/28 by multiplying top and bottom by 7 Rewrite x/7 as 4x/28 by multiplying top and bottom by 4 Write common bottom line and combine the top lines. Previous Next Back to Exercise M1 Menu

Exercise M1 Question 3 3. Write as a single expression:
4 and 5 have common multiple of 20 Rewrite 2x/5 as 4x/10 by multiplying top and bottom by 2 Write common bottom line and combine the top lines. Previous Next Back to Exercise M1 Menu

Exercise M1 Question 4 4. Write as a single expression:
4 and 5 have common multiple of 20 Rewrite 3x/4 as 15x/20 by multiplying top and bottom by 5 Rewrite 2x/5 as 8x/20 by multiplying top and bottom by 4 Write common bottom line and combine the top lines. 4. Write as a single expression: Previous Next Back to Exercise M1 Menu

Exercise M1 Question 5 5. Write as a single expression:
x and 5 multiply to give 5x Rewrite 4/x as 20/5x by multiplying top and bottom by 5 Rewrite x/5 as by multiplying top and bottom by x Write common bottom line and combine the top lines. Previous Next Back to Exercise M1 Menu

Exercise M1 Question 6 x2 and x have common multiple of x2
Rewrite as by multiplying top and bottom by x Write common bottom line and combine the top lines. 6. Write as a single expression: Previous Next Back to Exercise M1 Menu

Exercise M1 Question 7 7. Write as a single expression:
3 and x multiply to give 3x Rewrite x/3 as by multiplying top and bottom by x Rewrite 5/x as 15/x by multiplying top and bottom by 3 Write common bottom line and combine the top lines. Previous Next Back to Exercise M1 Menu

Exercise M1 Question 8 8. Write as a single expression:
4 and 8x both divide into 8x Rewrite x/4 as by multiplying top and bottom by 2x Write common bottom line and combine the top lines. 8. Write as a single expression: Previous Next Back to Exercise M1 Menu

Exercise M1 Question 9 9. Write as a single expression:
Common denominator 10p2 First term multiply top and bottom by 2 Second term multiply top and bottom by p2 Write common bottom line and combine the top lines. Previous Next Menu Back to Exercise M1 Menu

Exercise M2 Menu Simplifying rational expressions
Answer each question as a single expression. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Next Back to Merit Menu

Exercise M2 Question 1 1. Simplify fully: x2 – 3x = x(x – 3)
Cancel out the x Next Back to Exercise M2 Menu

Exercise M2 Question 2 2. Simplify fully: 3x + 6 = 3(x + 2) 3 = 3  1
Cancel out the 3 Previous Next Back to Exercise M2 Menu

Exercise M2 Question 3 3. Simplify fully: 2x2 – 4x = 2x(x – 2)
Cancel out the 2x Previous Next Back to Exercise M2 Menu

Exercise M2 Question 4 4. Simplify fully: x2 + 3x + 2 = (x + 1)(x + 2)
Cancel out the (x + 2) Previous Next Back to Exercise M2 Menu

Exercise M2 Question 5 5. Simplify fully: x – 4 = (x – 4)  1
x2 – 2x – 8 = (x – 4)(x + 2) Cancel out the (x – 4) Previous Next Back to Exercise M2 Menu

Exercise M2 Question 6 6. Simplify fully: x – 5 = (x – 5)  1
x2 + 3x – 40 = (x + 8)(x – 5) Cancel out the (x – 5) Previous Next Back to Exercise M2 Menu

Exercise M2 Question 7 7. Simplify fully: x2 – 16 = (x – 4)(x + 4)
Cancel out the (x – 4) Previous Next Back to Exercise M2 Menu

Exercise M2 Question 8 8. Simplify fully: x2 + 5x = x(x + 5)
5x3 + 25x2 = 5x (x)(x + 5) Cancel out the x(x + 5) Previous Next Back to Exercise M2 Menu

Exercise M2 Question 9 9. Simplify fully:
p2 + 2p – 15 = (p + 5)(p – 3) p2 – 9 = (p + 3)(p – 3) Cancel out the (p – 3) Previous Next Menu Back to Exercise M2 Menu

Solve the quadratic pattern in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Complete the table and write a rule for when n metres has been added. Write a rule for the number of triangles needed for Design n. Find a rule to give the number of plants needed if n rounds are to be added. How many tins would there be in x layers? Q1. Q2. Q3. Q4. Next Back to Merit Menu

Exercise M3 Question 1 3 gives 7  6 = 42 4 gives 8  7 = 56
1. A garden area is a rectangle 4 m by 3 m. It is being added to by extending 1 metre at a time along 2 adjacent sides. After 1 metre is added it has area 5 m by 4 m = 20 m2. After a second metre is added it has area 6 m by 5 m = 30 m2. Complete the table and write a rule for when n metres has been added. Length added 1 2 3 4 5 6 Area 5x4=20 6x5=30 7x6=42 3 gives 7  6 = 42 4 gives 8  7 = 56 5 gives 9  8 = 72 6 gives 10  9 = 90 n gives (n + 4)(n + 3) Answer Alternative Method Next Question Back to Exercise M3 Menu

Exercise M3 Q1 (Answer b) Question 1: Alternative method
Back to the question Length added, n Area Diff 2nd diff 1 20 10 2 30 12 3 42 14 4 56 16 5 72 Question 1: Alternative method Half second difference = 1, so rule is to do with (1) n2. n2 gives , 4, 9, 16, 25, … But Area is , 30, 42, 56, 72, … Need an extra 19, 26, 33, 40, 47,… This is a linear pattern with difference 7, so can be found by 7n + 12 Therefore rule is n2 + 7n + 12 Next Question Back to Exercise M3 Menu

Exercise M3 Question 2 2. A pattern is made from the basic design as shown. Write a rule for the number of triangles needed for Design n. One way of seeing this pattern is: Design 1 = 6 triangles (2 by 2 square doubled to get triangles – 2 corner triangles) = 2 x (2 x 2) – 2 x 1 Design 2 = 2 x (3 x 4) – 2 x (2 x 2) doubled rectangle – 2 triangles which form a square Design 3 = 2 x (4 x 6) – 2 x (3 x 3) Design 4 would have rectangle 5 by 8 = 2 x (5 x 8) – 2 x (4 x 4) Design n would be = 2(n + 1)(2n) – 2(n x n) = 4n2 + 4n – 2n2 = 2n2 + 4n << figure Ch01.12>> Answer Alternative Method Previous Next Question Back to Exercise M3 Menu

Exercise M3 Q2 (Answer b) Previous Next Question
Back to the question Design, n Triangles Diff 2nd diff 1 6 10 2 16 4 14 3 30 18 48 22 5 70 Question 2: Alternative method There are several different ways of seeing the design which lead to the same answer. This approach is the mathematical one: Half second difference = 2, so rule is to do with 2n2. n2 gives , 4, 9, 16, 25, … 2n2 gives , 8, 18, 32, 50 Triangles is , 16, 30, 48, 70, … Need an extra 4, 8, 12, 16, 20,… This is a linear pattern with difference 4, so can be found by 4n Therefore rule is: 2n2 + 4n Previous Next Question Back to Exercise M3 Menu

Exercise M3 Question 3 3. Plants are to be spaced out equally in a rectangular field. From a centre rectangular area with 8 plants, the plants are set out in rounds as shown in the diagram. Find a rule to give the number of plants needed if n rounds are to be added. 1 round 4 by 6 = 24 2 rounds 6 by 8 = 48 3 rounds 8 by 10 = 80 Dimensions are going up in 2s so linear patterns. n rounds (2n + 2) by (2n + 4) So (2n + 2)(2n + 4) plants Answer << figure Ch01.13>> Alternative Method Previous Next Question Back to Exercise M3 Menu

Exercise M3 Q3 (Answer b) Previous Next Question
Back to the question Rounds added, n Plants Diff 2nd diff 1 24 2 48 8 32 3 80 40 4 120 5 168 Question 3: Alternative method There are several different ways of seeing the design which lead to the same answer. This approach is the mathematical one: Half second difference = 4, so rule is to do with 4n2. n2 gives , 4, 9, 16, 25, … 4n2 gives , 16, 36, 64, 100 Triangles is , 48, 80, 120, 168, … Need an extra 20, 32, 44, 56, 68,… This is a linear pattern with difference 12, so can be found by 12n + 8 Therefore rule is: 4n2 + 12n + 8 Previous Next Question Back to Exercise M3 Menu

Exercise M3 Question 4 4. Cans are stacked on a shelf as shown in the diagrams. There are 4 cans in 1 layer, 11 cans if there are 2 layers, 21 cans if there are 3 layers and so on. How many cans would there be in x layers? 1 layer = 4 cans 2 layers = 0.5 x (2 x 11) 3 layers = 0.5 x (3 x 14) 4 layers = 0.5 x (4 x 17) Linear pattern for values in brackets, so x layers has 0.5 x x x (3x+5) =0.5x(3x+5) Answer << figure Ch01.14>> Alternative Method Previous Next Menu Back to Exercise M3 Menu

Back to the question Layers, x Cans Diff 2nd diff 1 4 7 2 11 3 10 21 13 34 16 5 50 Question 4: Alternative method There are several different ways of seeing the design which lead to the same answer. This approach is the mathematical one: Half second difference = 3, so rule is to do with 1.5x2. x2 gives , 4, 9, 16, 25, … 1.5x2 gives , 6, 13.5, 24, 37.5 Cans is , 11, 21, 34, 50, … Need an extra 2.5, 5, 7.5, 10, 12.5,… This is a linear pattern with difference 2.5, so can be found by 2.5x Therefore rule is: 1.5x x Previous Next Menu Back to Exercise M3 Menu

Exercise M4 Menu Change the subject
Make k the subject in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Q1. x = 15k + 7 m = k2 – 5 z = Q2. P = 3(k – 2) B = g = 4a2k – 2p Q3. Q4. Q5. Q6. Next Back to Merit Menu

Exercise M4 Question 1 1. Make k the subject in: x = 15k + 7
15k + 7 = x – – 7 15k = x – 7   15 k = (x – 7)  15 or Previous Next Back to Exercise M4 Menu

Exercise M4 Question 2 2. Make k the subject in: P = 3(k – 2)
3(k – 2) = P   3 k – 2 = P/3 k = P/ or Previous Next Back to Exercise M4 Menu

Exercise M4 Question 3 3. Make k the subject in: m = k2 – 5 k2 – 5 = m
k2 = m + 5 take square root k = Previous Next Back to Exercise M4 Menu

Exercise M4 Question 4 4. Make k the subject in B = x 3 x 3 3B = k – 5
k – 5 = 3B k = 3B + 5 Previous Next Back to Exercise M4 Menu

Exercise M4 Question 5 k = 4 5. Make k the subject in z = + q k
– q – q z – q = 4/k x k x k k(z – q) = 4 (z – q) (z – q) k = 4 z – q Previous Next Back to Exercise M4 Menu

Exercise M4 Question 6 6. Make k the subject in g = 4a2k – 2p
4a2k – 2p = g + 2p + 2p 4a2k = g + 2p  4a2  4a2 k = g + 2p 4a2 Previous Next Menu Back to Exercise M4 Menu

Solve the inequations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. 4x – 8 < 20 4(2 – x) > 16 3x + 8 > 5x Q1. Q2. Q3. Q4. Q5. Q6. Next Back to Merit Menu

Exercise M5 Question 1 1. Solve: 4x – 8 < 20 + 8 +8 4x < 28
4x < 28   4 x < 7 Next Back to Exercise M5 Menu

Exercise M5 Question 2 2. Solve: 3x + 8 > 5x
Swap sides, reversing inequation 5x < 3x + 8 – 3x – 3x 2x < 8  2  2 x < 4 Previous Next Back to Exercise M5 Menu

Exercise M5 Question 3  4  4 x < -2 3. Solve: 4(2 – x) > 16
  4 2 – x > 4 – – 2 -x > 2 -x  -1 < 2  Note: Change of sign direction x < -2 Previous Next Back to Exercise M5 Menu

Exercise M5 Question 4 x 3 x 3 k – 5 < 24 + 5 + 5 k < 29
4. Solve: x x 3 k – 5 < 24 k < 29 Previous Next Back to Exercise M5 Menu

Exercise M5 Question 5 + 3 + 3 k/5 > 14 x 5 x 5 k > 70 5. Solve:
k/5 > 14 x x 5 k > 70 Previous Next Back to Exercise M5 Menu

Exercise M5 Question 6 6. Solve: – 6 – 6 -k/6 > 5 x 6 x 6
– – 6 -k/6 > 5 x x 6 -k > 30 -k  < 30  -1 Note: Change of sign direction k < -30 Previous Next Menu Back to Exercise M5 Menu

Exercise M6 Menu Form and solve equations and inequations
Click on the buttons to go to each problem or scroll to each question/answer using the ‘next’ arrow (below). Write an equation or inequation to represent each problem, then solve it. Press the answer button for the correct answer. Click to move from step to step in the solution. Five more than three times a number is forty seven. What is the number? How many items of jewellery were sold on commission? How many weeks will it be before the account is overdrawn? What day will Marama be jogging more than 5 km? After how many weeks will Tyrone have more than \$500? What is the mechanic charging per hour for labour? How many calls can be made, without spending more than \$60 in a week? How many glasses of drink can be poured? Q1. Q2. Q4. Q3. Q5. Q6. Q7. Q8. Next Back to Merit Menu

Exercise M6 Question 1 3x + 5 = 47
1. Write an equation or inequation to represent: Five more than three times a number is forty seven. What is the number? 3x + 5 = 47 – – 5 3x = 42   3 x = 14 Answer Next Back to Exercise M6 Menu

Exercise M6 Question 2 16w + 83 > 500 w = number of weeks – 83 – 83
2. Write an equation or inequation to represent: Tyrone saves \$16 a week for a holiday. If he starts with \$83, after how many weeks will he have more than \$500? 16w > w = number of weeks – – 83 16w > 417   16 w > Therefore 27 weeks Answer Previous Next Back to Exercise M6 Menu

Exercise M6 Question 3 3. Write an equation or inequation to represent: Trish sells jewellery on commission at a jewellery evening. She gets \$8 for each item she sells, but it costs her \$45 to cover the costs of the jewellery evening. If she makes \$203 one evening, how many items of jewellery did she sell? 8j = j = number of jewellery items 8j = 248  8  8 j = 31 Therefore 31 items Answer Previous Next Back to Exercise M6 Menu

Exercise M6 Question 4 3.5h + 56 = 248.50 h = hourly rate – 56 – 56
4. Write an equation or inequation to represent: A mechanic charges \$56 for parts and has worked for 3.5 hours on the car. If the total account was \$248.50, what is the mechanic charging per hour for labour? 3.5h = h = hourly rate – – 56 3.5h =   3.5 h = 55 Therefore \$55 per hr Answer Previous Next Back to Exercise M6 Menu

Exercise M6 Question 5 5. Write an equation or inequation to represent: Jenny has \$1364 in an account. If she pays her rent from this account and it is \$75 per week, how many weeks will it be before her account is overdrawn? 1364 – 75w < w = number of weeks – – 1364 -75w < -75w  >  -75 w > Therefore 19 weeks Answer Previous Next Back to Exercise M6 Menu

Exercise M6 Question 6 6. Write an equation or inequation to represent: Steven pays \$20 a month for the telephone account. Phone calls to his girlfriend in another town cost \$2.75 each. He wants to call his girlfriend as much as possible. How many calls can he make but not spend more than \$60 in a week? c < c = number of calls – – 20 2.75c < 40   2.75 c < Therefore 14 calls Answer Previous Next Back to Exercise M6 Menu

Exercise M6 Question 7 7. Write an equation or inequation to represent: Marama is increasing the distance she jogs by one more telephone pole length, which means she adds 24 m each day she jogs. If on day 1 she jogged 2 km (2000 m), on what day’s jogging will she be jogging more than 5 km? (d – 1) > d = number of days d > 5000 – – 1976 24d > 3024   24 d > 126 Therefore Day 127 Answer Previous Next Back to Exercise M6 Menu

Exercise M6 Question 8 8. Write an equation or inequation to represent: From a 9.5 L (9500 ml) drink container, 150 ml glasses are being poured. If 30 ml was spilt from the container, how many glasses of drink could be poured? 150g = g = number of glasses Note: Could be < instead of = – – 30 150g = 9470   150 g = Therefore 63 glasses Answer Previous Next Menu Back to Exercise M6 Menu

Solve the simple quadratic equations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. Q1. x2 + 6x + 5 = 0 x2 + 4x – 12 = 0 x2 + 6x = 16 x2 – 3x = 10 Q2. x2 + 9x + 18 = 0 x2 – 5x – 24 = 0 x2 + 10x = -24 x2 – 2x = 8 Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Merit Menu

Exercise M7 Question 1 Pairs of factors of 5 are 1, 5; -1, -5
1. Solve: x2 + 6x + 5 = 0 Pairs of factors of 5 are 1, 5; -1, -5 Pair that add to give 6 are 1, 5 Therefore (x + 1)(x + 5) = 0 x + 1 = 0 or x + 5 = 0 – 1 – – 5 – 5 x = or x = -5 Next Back to Exercise M7 Menu

Exercise M7 Question 2 2. Solve: x2 + 9x + 18 = 0
Pairs of factors of 18 are: 1, 18; 2, 9; 3, 6; -1, -18; -2, -9; -3, -6 Pair that add to 9 are 3, 6 Therefore (x + 3)(x + 6) = 0 x + 3 = 0 or x + 6 = 0 – 3 – – 6 – 6 x = or x = -6 Previous Next Back to Exercise M7 Menu

Exercise M7 Question 3 3. Solve: x2 + 4x – 12 = 0
Pairs of factors of -12 are: 1, -12; 2, -6; 3, -4; 4, -3; 6, -2; 12, -1 Pair that add to 4 are 6, -2 Therefore (x + 6)(x – 2) = 0 x + 6 = or x – 2 = 0 – 6 – x = or x = 2 Previous Next Back to Exercise M7 Menu

Exercise M7 Question 4 4. Solve: x2 – 5x – 24 = 0
Pairs of factors of -24 are: 1, -24; 2, -12; 3, -8; 4, -6; 6, -4; 8, -3; 12, -2; 24, -1 Pair that add to -5 are 3, -8 Therefore (x + 3)(x – 8) = 0 x + 3 = or x – 8 = 0 – 3 – x = or x = 8 Previous Next Back to Exercise M7 Menu

Exercise M7 Question 5 5. Solve: x2 + 6x = 16
Get all terms on left by – giving x2 + 6x – 16 = 0 Pairs of factors of -16 are: 1, -16; 2, -8; 4, -4; 8, -2; 16, -1 Pair that add to 6 are 8, -2 Therefore (x + 8)(x – 2) = 0 x + 8 = or x – 2 = 0 – 8 – x = or x = 2 Previous Next Back to Exercise M7 Menu

Exercise M7 Question 6 6. Solve: x2 + 10x = -24
Get all terms on left by + 24 giving x2 + 10x + 24 = 0 Pairs of factors of 24 include: 1, 24; 2, 12; 4, 6 Pair that add to 10 are 4, 6 Therefore (x + 4)(x + 6) = 0 x + 4 = or x + 6 = 0 – 4 – – 6 – 6 x = or x = -6 Previous Next Back to Exercise M7 Menu

Exercise M7 Question 7 7. Solve: x2 – 3x = 10
Get all terms on left by – 10 giving x2 – 3x – 10 = 0 Pairs of factors of -10 are: 1, -10; 2, -5; 5, -2; 10, -1 Pair that add to -3 are 2, -5 Therefore (x + 2)(x – 5) = 0 x + 2 = or x – 5 = 0 – 2 – x = or x = 5 Previous Next Back to Exercise M7 Menu

Exercise M7 Question 8 8. Solve: x2 – 2x = 8
Get all terms on left by – 8 giving x2 – 6x – 8 = 0 Pairs of factors of -8 are: 1, -8; 2, -4; 4, -2; 8, -1 Pair that add to -2 are 2, -4 Therefore (x + 2)(x – 4) = 0 x + 2 = or x – 4 = 0 – 2 – x = or x = 4 Previous Next Menu Back to Exercise M7 Menu

Solve the quadratic equations in each question. Press the button for the correct answer or scroll to each question/answer using the ‘next’ arrow (below). Click to move from step to step in the solution. 2x2 + 6x – 56 = 0 4x2 + 4x = 288 5p2 + 32p + 12 = 0 4x2 + 20x + 25 = 0 Q1. Q2. 3x2 + 12x + 9 = 0 3x2 – 19x + 20 = 0 4x2 – 7x + 3 = 0 3x2 + 19x = 14 Q3. Q4. Q5. Q6. Q7. Q8. Next Back to Merit Menu

Exercise M8 Question 1 Common factor of 2, so divide by this leaving
1. Solve: 2x2 + 6x – 56 = 0 Common factor of 2, so divide by this leaving x2 + 3x – 28 = 0 Factors of -28 are: 1, -28; 2, -14; 4, -7; 7, -4; 14, -2; 28, -1 Pair that add to give 3 are 7, -4 Therefore (x + 7)(x – 4) = 0 x + 7 = or x – 4 = 0 – 7 – x = or x = 4 Next Back to Exercise M8 Menu

Exercise M8 Question 2 2. Solve: 3x2 + 12x + 9 = 0
Common factor of 3, so divide by this leaving x2 + 4x + 3 = 0 Factors of 3 are 1, 3; -1, -3; Pair that add to give 4 are 1, 3 Therefore (x + 1)(x + 3) = 0 x + 1 = or x + 3 = 0 – 1 – – 3 – 3 x = or x = -3 Previous Next Back to Exercise M8 Menu

Exercise M8 Question 3 3. Solve: 4x2 + 4x = 288
Common factor of 4, so divide by this leaving x2 + x = 72 Subtract 72 from both sides, so x2 + x – 72 = 0 Factors of -72 include 4, -18; 6, -12; 8, -9; 9, -8; Pair that add to give 1 are: 9, -8 Therefore (x + 9)(x – 8) = 0 x + 9 = or x – 8 = 0 – 9 – x = or x = 8 Previous Next Back to Exercise M8 Menu

Exercise M8 Question 4 4. Solve: 3x2 – 19x + 20 = 0 No common factor;
x2 term and constant multiplied gives 60x2 Pairs that multiply to give 60x2 include: -x, -60x; -2x, -30x; -3x,-10x; -4x, -15x; -5x, -12x; -6x, -10x Pair that add to give -19x are -4x and -15x 3x2 – 4x – 15x + 20 = 0 x(3x – 4) – 5(3x – 4) = 0 Therefore (3x – 4)(x – 5) = 0 3x – 4 = or x – 5 = 0 3x = x = 4/ or x = 5 Previous Next Back to Exercise M8 Menu

Exercise M8 Question 5 5. Solve: 5p2 + 32p + 12 = 0
No common factor; x2 term and constant multiplied gives 60p2 Pairs that multiply to give 60x2 include: p, 60p; 2p, 30p; Pair that add to give to 32p are 2p and 30p 5p2 + 2p + 30p + 12 = 0 p(5p + 2) + 6(5p + 2) = 0 Therefore (5p + 2)(p + 6) = 0 5p + 2 = or p + 6 = 0 5p = – 6 – 6 p = -2/5 or x = -6 Previous Next Back to Exercise M8 Menu

Exercise M8 Question 6 6. Solve: 4x2 – 7x + 3 = 0 No common factor;
x2 term and constant multiplied gives 12x2 Pairs that multiply to give 12x2 include: -x, -12x; -2x, -6x; -3x, -4x; Pair that add to give to -7x are -3x and -4x 4x2 – 3x – 4x + 3 = 0 x(4x – 3) – 1(4x – -3) = 0 Therefore (4x – 3)(x – 1) = 0 4x – 3 = or x – 1 = 0 4x = x = 3/ or x = 1 Previous Next Back to Exercise M8 Menu

Exercise M8 Question 7 7. Solve: 4x2 + 20x + 25 = 0
No common factor; x2 term and constant multiplied gives 100x2 Pairs that multiply to give 100x2 include: 2x, 50x; 4x, 25x; 5x, 20x; 10x, 10x; Pair that add to give 20x are 10x and 10x 4x2 + 10x + 10x + 25 = 0 2x(2x + 5) + 5(2x + 5) = 0 Therefore (2x + 5)(2x + 5) = 0 2x + 5 = Note: Other factor gives same result 2x = -5 x = -5/2 or Previous Next Back to Exercise M8 Menu

Exercise M8 Question 8 8. Solve: 3x2 + 19x = 14 3x2 + 19x – 14 = 0
No common factor; take 14 from both sides 3x2 + 19x – 14 = 0 x2 term and constant multiplied gives -42x2 Pairs that multiply to give -42x2 include: 7x, -6x; 14x, -3x; 21x,-2x; Pair that add to give 19x are 21x and -2x 3x2 + 21x – 2x – 14 = 0 3x(x + 7) – 2(x + 7) = 0 Therefore (x + 7)(3x – 2) = 0 x + 7 = or x – 2 = 0 – – x = 2 x = or x = 2/3 Previous Next Menu Back to Exercise M8 Menu

Exercise M9 Menu Solve simultaneous equations using elimination
Solve these equations. Press the button for the correct answer or scroll using the ‘next’ arrow (below). Click to move through the solution. 2x + 7y = 68 12x – 7y = 16 4x + 9y = 11 -2x – 3y = -1 5f + 6g = 26 f +   g = 4 4k + 7n = 41 3k – 2n = 9 3t + 4u = 25 5t – 6u = 11.9 3x – 4y = 20 -3x + 20y = 20 7x y = 11 4x – 10y = 32 100d – 20e = 114 5d + 4e = 17.2 34p + 11q = 91 4p – 2q = 14 2x + 8y = 13 5x – 7y = -62 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10. Next Back to Merit Menu

Exercise M9 Question 1 1. Solve: 2x + 7y = 68 12x – 7y = 16
As 7y and -7y will cancel out, add the two equations to get: 14x = 84 gives x = 6 Substitute into first equation to find y y = 68 gives 7y = 56 so y = 8 Check second equation 12  6 – 7  8 = 72 – 56 = 16 x = 6, y = 8 Next Back to Exercise M9 Menu

Exercise M9 Question 2 2. Solve: 3x – 4y = 20 -3x + 20y = 20
As 3x and -3x will cancel out, add the two equations to get: 16y = 40 gives y = 2.5 Substitute into first equation to find x 3x – 10 = 20 gives 3x = 30 so x = 10 Check second equation -3   2.5 = = 20 x = 10, y = 2.5 Previous Next Back to Exercise M9 Menu

(Note: could have been by 2 to eliminate the x with 4x and -4x)
Exercise M9 Question 3 3. Solve: 4x + 9y = 11 -2x – 3y = -1 Need to multiply equation 2 by 3 so 9y and -9y will cancel out (Note: could have been by 2 to eliminate the x with 4x and -4x) -6x – 9y = -3 Add this new equation to the first equation -2x = 8 gives x = -4 Substitute into first equation to find y y = 11 gives 9y = 27 so y = 3 Check second equation -2  -4 – 3  3 = 8 – 9 = -1 x = -4, y = 3 Previous Next Back to Exercise M9 Menu

Exercise M9 Question 4 4. Solve: 7x + 5y = 11 4x – 10y = 32
Need to multiply equation 1 by 2 so 10y and -10y will cancel out 14x y = 22 Add this new equation to the second equation 18x = 54 gives x = 3 Substitute into first equation to find y y = 11 gives 5y = -10 so y = -2 Check second equation 4  3 – 10  -2 = 12 – -20 = 32 x = 3, y = -2 Previous Next Back to Exercise M9 Menu

Exercise M9 Question 5 5. Solve: 5f + 6g = 26 f + g = 4
Need to multiply equation 2 by -5 so 5f and -5f will cancel out (Note could have been by -6 to eliminate using 6g and -6g) -5f g = -20 Add this new equation to the second equation gives g = 6 Substitute into first equation to find y 5f = 26 gives 5f = -10 so f = -2 Check second equation = 4 f = -2, g = 6 Previous Next Back to Exercise M9 Menu

Exercise M9 Question 6 6. Solve: 100d – 20e = 114 5d + 4e = 17.2
Multiply equation 2 by 5 so -20e and 20e will cancel out (Note could have been by -20 to eliminate using 100d and -100d) 25d e = 86 Add this new equation to the first equation 125d = 200 gives d = 1.6 Substitute into first equation to find e 160 – 20e = 114 gives -20e = so e = 2.3 Check second equation 5   2.3 = = 17.2 d = 1.6, e = 2.3 Previous Next Back to Exercise M9 Menu

Exercise M9 Question 7 7. Solve: 4k + 7n = 41 3k – 2n = 9
Multiply eqn 1 by 2 and eqn 2 by 7 so 14n and -14n will cancel out. (Note could have been eqn 1 by 3 and eqn 2 by -4 to eliminate using 12k and -12k) 8k n = 82 21k – 14n = 63 Add these k = 145 gives k = 5 Substitute into first equation to find n n = 41 gives 7n = 21 so n = 3 Check second equation 3  5 – 2  3 = 15 – 6 = 9 k = 5, n = 3 Previous Next Back to Exercise M9 Menu

Exercise M9 Question 8 8. Solve: 34p + 11q = 91 4p – 2q = 14
Multiply eqn 1 by 2 and eqn 2 by 11 so 22q and -22q will cancel out. (Note could have been eqn 1 by 2 and eqn 2 by -17 to eliminate using 68p and -68p) 68p q = 182 44p – 22q = 154 Add these 112p = 336 gives p = 3 Substitute into first equation to find q q = 91 gives 11q = -11 so q = -1 Check second equation 4  3 – 2  -1 = 12 – -2 =14 p = 3, q = -1 Previous Next Back to Exercise M9 Menu

Exercise M9 Question 9 9. Solve: 3t + 4u = 25 5t – 6u = 11.9
Multiply eqn 1 by 3 and eqn 2 by 2 so 12u and -12u will cancel out. (Note could have been eqn 1 by 5 and eqn 2 by -3 to eliminate using 15t and -15t) 9t u = 75 10t – 12u = 23.8 Add these 19t = 98.8 gives t = 5.2 Substitute into first equation to find u u = 25 gives 4u = 9.4 so u = 2.35 Check second equation 5  5.2 – 6  2.35 = 26 – 14.1 = 11.9 t = 5.2, u = 2.35 Previous Next Back to Exercise M9 Menu

Exercise M9 Question 10 10. Solve: 2x + 8y = 13 5x – 7y = -62
Multiply eqn 1 by 7 and eqn 2 by 8 so 56y and -56y will cancel out. (Note could have been eqn 1 by 5 and eqn 2 by -2 to eliminate using 10x and -10x) 14x y = 91 40x – 56y = -496 Add these 54x = -405 gives x = -7.5 Substitute into first equation to find y y = 13 gives 8y = 28 so y = 3.5 Check second equation 5  -7.5 – 7  3.5 = – 24.5 = -62 x = -7.5, y = 3.5 Previous Next Menu Back to Exercise M9 Menu

Exercise M10 Menu Solve simultaneous equations using substitution
Solve these equations. Press the button for the correct answer or scroll using the ‘next’ arrow (below). Click to move from step to step in the solution. y = 2x + 8 3x + y = 16 x = y – 11 2x – 3y = -1 f = 2g + 6 f + g = 4 k = 20 – n 2n – 3k = 8 y = 3x + 17 y = 2x – 3 x = y – 20 x + 3y = 40 y = x 4x – 2y = 15 d = 5e – 17 2d + 6e = 18 q = 7 – 6p 4p – 2q = 14 y = -3x – 7 4x – 2y = -12 Q1. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Q9. Q10. Next Back to Merit Menu

Exercise M10 Question 1 1. Solve: y = 2x + 8 3x + y = 16
Replace y in equation 2 by what it equals in equation 1: 3x + (2x + 8) = 16 Simplify 5x = 16 gives 5x = 8 so x = 1.6 Substitute into first equation to find y y = 2  so y = 11.2 Check second equation 3  = = 16 x = 1.6, y = 11.2 Next Back to Exercise M10 Menu

Exercise M10 Question 2 2. Solve: x = y – 20 x + 3y = 40
Replace x in equation 2 by what it equals in equation 1: (y – 20) + 3 y = 40 Simplify 4 y – 20 = 40 gives y = 60 so y = 15 Substitute into first equation to find x x = 15 – 20 so x = -5 Check second equation  15 = = 40 x = -5, y = 15 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 3 3. Solve: x = y – 11 2x – 3y = -1
Replace x in equation 2 by what it equals in equation 1: 2(y – 11) – 3y = -1 Simplify -y – 22 = -1 gives y = 21 so y = -21 Substitute into first equation to find x x = -21 – 11 so x = -32 Check second equation 2  -32 – 3  -21 = = -1 x = -32, y = -21 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 4 4. Solve: y = 7 + 3x 4x – 2y = 15
Replace y in equation 2 by what it equals in equation 1: 4x – 2(7 + 3x) = 15 Simplify -2x – 14 = 15 gives x = 29 so x = -14.5 Substitute into first equation to find y y = so y = -36.5 Check second equation 4  – 2  = = 15 x = -14.5, y = -36.5 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 5 5. Solve: f = 2g + 6 f + g = 24
Replace f in equation 2 by what it equals in equation 1: (2g + 6) + g = 24 Simplify 3g + 6 = 24 gives g = 18 so g = 6 Substitute into first equation to find f f = 2  6 + 6 so f = 18 Check second equation = 24 f = 18, g = 6 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 6 6. Solve: d = 5e – 17 2d + 6e = 18
Replace d in equation 2 by what it equals in equation 1: 2(5e – 17) + 6e = 18 Simplify 16e – 34 = 18 gives e = 52 so e = 3.25 Substitute into first equation to find d d = 5  3.25 – 17 so d = -0.75 Check second equation 2   3.25 = = 18 d = -0.75, e = 3.25 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 7 7. Solve: k = 20 – n 2n – 3k = 8
Replace k in equation 2 by what it equals in equation 1: 2n – 3(20 – n) = 8 Simplify 5n – 60 = 8 Note: take care with the – (..-..) to get + gives n = 68 so n = 13.6 Substitute into first equation to find k k = 20 – 13.6 so k = 6.4 Check second equation 2  13.6 – 3  6.4 = 27.2 – 19.2 = 8 k = 6.4, n = 13.6 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 8 8. Solve: q = 7 – 6p 4p – 2q = 14
Replace q in equation 2 by what it equals in equation 1: 4p – 2(7 – 6p) = 14 Simplify 16p – 14 = 14 Note: take care with the – (..-..) to get + gives p = 28 so p = 1.75 Substitute into first equation to find q q = 7 – 6  1.75 so q = 7 – 10.5 = -3.5 Check second equation 4  1.75 – 2  -3.5 = = 14 p = 1.75, q = -3.5 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 9 9. Solve: y = 3x + 17 y = 2x – 3
Replace y in equation 2 by what it equals in equation 1: 3x = 2x – 3 Simplify: – 17 from both sides 3x = 2x – 20 -2x gives x = Substitute into first equation to find y y = 3  so y = = -43 Check second equation y = 2  -20 – 3 = as required x = -20, y = -43 Previous Next Back to Exercise M10 Menu

Exercise M10 Question 10 10. Solve: y = -3x – 7 4x – 2y = -12
Replace y in equation 2 by what it equals in equation 1: 4x – 2(-3x – 7) = -12 Simplify 10x = -12 10x = -26 x = -2.6 Substitute into first equation to find y y = -3  -2.6 – 7 so y = 7.8 – 7 = 0.8 Check second equation 4  -2.6 – 2  0.8 = – 1.6 = -12 x = -2.6, y = 0.8 Previous Next Menu Back to Exercise M10 Menu

Exercise M11 Menu Solve simultaneous equations in context
Solve each of the simultaneous equations. Press the button for question details or scroll to each question using the ‘next question’ arrow (below). Click to move from step to step in the solution. Q1. Find the price of a chicken pie. Find the price of the drink and chips. Find how many metres of wallpaper. Find the number of 2-legged pets. Find the adult price to swim at the pool. Find the time taken to prepare each sandwich. Find the size of the larger number. Find how many trees Ricky pruned. Q2. Q3. Q4. Q5. Q6. Q7. Q8. Next Question Back to Merit Menu

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Exercise M11 Question 1 1. Chicken pies cost 50c more than mince pies. If 6 mince and 9 chicken pies cost a total of \$40.50, solve the pair of simultaneous equations: c = m + 0.5 6m + 9c = 40.5 to find the price of a chicken pie. Answer Next Question Back to the Exercise 11 Menu

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Exercise M11 Q1 Answer Question: Find the cost of a chicken pie Answer: c = m + 0.5 6m + 9c = 40.5 Substitute for c in the second equation: 6m + 9(m + 0.5) = 40.5 Simplifying: m = 40.5 Giving: m = 36 m = 2.4 Substitute into the first equation: c = m + 0.5 c = = 2.9 Read the problem and check the words for chicken pies at \$2.90 and mince pies \$2.40 Chicken pies 50c more than mince, 6 mince and 9 chicken pies = 6  \$  \$2.90 = \$40.50 Chicken pie costs \$2.90 Back to the question Next Question Back to the Exercise 11 Menu

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Exercise M11 Question 2 2. A family with 2 adults and 3 children pay \$14.80 to go swimming at the pool. Another group with 5 adults and 2 children pay \$30.40 in total to swim at the pool. Solve the simultaneous equations 2a + 3c = 14.8 5a + 2c = 30.4 to find the price for an adult to go swimming at the pool. Answer Previous Next Question Back to the Exercise 11 Menu

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Exercise M11 Q2 Answer Back to the question Question: Find the adult price at a swimming pool. Answer: 2a + 3c = 14.8 5a + 2c = 30.4 Multiply first equation by 2: 4a + 6c = 29.6 Multiply second equation by -3: -15a – 6c = Adding gives: a = Dividing by -11 gives: a = 5.6 This gives adult price \$5.60, but best to check out by: Subsitute into the first equation c = 14.8 Giving: c = 3.6 so c = 1.2 Check wording of problem with 2 adults, 3 children \$14.80, 5 adults and 2 children (5  \$  \$1.20) = \$30.40 Adults price is \$5.60 Next Question Back to the Exercise 11 Menu

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Exercise M11 Question 3 3. Six bags of chips and 4 bottles of drink cost \$ The money needed for the bottle of drink is 10 cents less than twice the price of the chips. Solve the simultaneous equations 6c + 4d = 24.8 d = 2c – 0.1 and use the results to find the price of 1 bottle of drink and 1 bag of chips. Answer Previous Next Question Back to the Exercise 11 Menu

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Exercise M11 Q3 Answer Question: Find the price of 1 bottle of drink and 1 bag of chips. Answer: 6c + 4d = 24.8 d = 2c – 0.1 Substitute for d in the first equation: 6c + 4(2c – 0.1) = 24.8 Simplifying: c – = 24.8 14c = c = 1.8 Substitute into the second equation: d = 2  1.8 – 0.1 d = 3.5 Read the problem and check the words: 6 chips and 4 drinks = 6  \$  \$3.50 = \$24.80 Drink (\$3.50) is 10c less than twice chips Price of 1 drink and 1 chips = d + c = \$ \$1.80 Price is \$5.30 Back to the question Next Question Back to the Exercise 11 Menu

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Exercise M11 Question 4 4. A kitchen worker takes 17.6 minutes to prepare 10 sandwiches and 2 burgers. She takes 14.6 minutes to prepare 4 sandwiches and 5 burgers. Assuming the worker is working at a constant rate, solve the equations: 10x + 2y = 17.6 4x + 5y = 14.6 to find the time taken to prepare each sandwich. Answer Previous Next Question Back to the Exercise 11 Menu

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Exercise M11 Q4 Answer Question: Find the time taken to prepare each sandwich. Answer: 10x + 2y = 17.6 4x + 5y = 14.6 x = time for sandwich, y = time for burger Want x, so eliminate y by multiplying first equation by 5: 50x + 10y = 88 Multiply second equation by -2: -8x – 10y = Adding gives: x = Dividing by 42: x = 1.4 This gives time for sandwich 1.4 min, but best to check out by: Substitute into the first equation y = 17.6 giving: 2y = 3.6 so y = 1.8 Check wording of problem with 10 sandwiches, 2 burgers 17.6 min, 4 sandwiches and 5 burgers (4   1.8) = 14.6 Time to prepare sandwich = 1.4 min Back to the question Next Question Back to the Exercise 11 Menu

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Exercise M11 Question 5 5. A formula is used to work out how many metres of wallpaper is needed for a standard room, using the distance around the room and the total width of the windows. The formula gives m for a room with perimeter 28 m and with windows with total width 3.6 m. The formula gives 71.1 m for a room with perimeter 16 m and windows with total length 1.5 m. Solve the equations: 28x + 3.6y = 16x + 1.5y = 71.1 to find out how many metres of wallpaper is allowed per metre of perimeter and per metre of total window width. Answer Previous Next Question Back to the Exercise 11 Menu

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Exercise M11 Q5 Answer Question: Find out how many metres of wallpaper. Answer: 28x + 3.6y = 16x + 1.5y = 71.1 x = perimeter of room, y = width of window Eliminate y by multiplying first equation by 1.5: 42x + 5.4y = Multiply second equation by -3.6: -57.6x – 5.4y = Adding: x = Dividing by -15.6: x = 4.5 Substitute into first equation y = giving: 3.6y = so y = -0.6 Check wording of problem: with 28 m perimeter, 3.6 m window = , 16 m perimeter, 1.5 m windows (16   -0.6) = 71.1 Formula allows 4.5 m per metre perimeter and takes 0.6 m per window width metre. Back to the question Next Question Back to the Exercise 11 Menu

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Exercise M11 Question 6 6. The sum of two numbers is 98 and the larger number is five more than twice the smaller number. Solve the equations: x + y = 98 y = 2x + 5 to find the size of the larger number. Answer Previous Next Question Back to the Exercise 11 Menu

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Exercise M11 Q6 Answer Question: Find the size of the larger number. Answer: x + y = 98 y = 2x + 5 Substitute for y in the first equation: x + (2x + 5) = 98 Simplifying x + 5 = 98 Giving x = 93 x = 31 Substitute into the second equation: y = 2  y = 67 Read the problem and check the words Sum of 31 and 67 is 98 67 is twice 31 plus 5 Question asks for larger number Larger number is 67 Back to the question Next Question Back to the Exercise 11 Menu

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Exercise M11 Question 7 7. There are 48 pets in a pet parade. The pets either have 4 legs or 2 legs. If there were 150 legs on the pets, solve the equations: x + y = 48 4x + 2y = 150 to find how many 2-legged pets are in the parade. Answer Previous Next Question Back to the Exercise 11 Menu

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Exercise M11 Q7 Answer Question: Find the number of 2-legged pets. Answer: x + y = 48 4x + 2y = 150 y = number of 2-legged pets. Multiply the first equation by -4 -4x – 4y = -192 Adding this to the second equation gives -2y = -42 y = 21 As a check, substitute into the first equation, x + 21 = 48 so x = 27 Check legs: 4   21 = = 150 Number of 2-legged pets is 21 Back to the question Next Question Back to the Exercise 11 Menu

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Exercise M11 Question 8 8. If Ricky had pruned 6 more trees he would have pruned twice as many as Steve. Together they pruned 96 trees. Solve the equations: x + 6 = 2y x + y = 96 to find how many trees Ricky pruned. Answer Previous Excellence Menu Back to the Exercise 11 Menu

Exercise M11 Q8 Answer Question: Find how many trees Ricky pruned.
Back to the question Question: Find how many trees Ricky pruned. Answer: x = 2y can be rearranged to give x – 2 y = -6 (1) We also know that x + y = 96 (2) x = number of trees Ricky pruned Multiplying equation (2) by 2 gives 2x + 2y = (3), Adding equation (3) to equation (1) gives 3x = 186 Therefore x, the number of trees Ricky pruned, = 62 As a check, substitute into the first equation, = 2y Therefore y = 34 This also fits the second equation x + y = 96 Back to Main Menu Excellence Menu Back to Merit Menu

Excellence Menu: Exercise E1 Solve equations in context by modelling
Solve each of the equations. Press the button for question details or scroll to each question using the ‘next question’ arrow (below). Click to move from step to step in the solution. How many movies were shown each week 5 years ago? Give the dimensions of each of the rectangular regions. Q1. Q2. Estimate the charge to erect a fence 50 m long. Give the maximum length of the garden. Q3. Q4. Calculate the worth when the cabinet is full. Give the side lengths of the original field. Q5. Q6. Calculate is the area of the section. Calculate how many tile layers there are. Q7. Q8. How much do they each earn? Q9. Q10. Calculate how many weeks. Next Question Back to Main Menu

Exercise E1 Question 1 1. The average number of movies shown in a movie complex each week this year has increased by 8 since 5 years ago. The average number of viewers per movie was 4 times the number of weekly movies shown 5 years ago, but is now half that number. If there are now 480 viewers per week, how many movies were shown each week 5 years ago? Answer Next Question Back to Excellence Menu

Exercise E1 Answer 1 1. How many movies were shown each week 5 years ago? Let number of movies per week 5 years ago = x. The number of movies shown each week has increased by 8 since 5 years ago. Number of movies per week now = x + 8 The average number of viewers per movie was 4 times the number of weekly movies 5 years ago so: Average number of viewers per movie was 4x but is now half that number so: Average number of viewers per movie is 2x There are now 480 viewers per week Number of viewers per week now = average number per movie x number of movies per week = 2x(x + 8) 2x(x + 8) = 480 2x2 + 16x – 480 = 0 Dividing by 2 x2 + 8x – 240 = 0 Factors of -400 include: -10, 24; -12, 20 (x – 12)(x + 20) = 0 x = -20 is not realistic. Therefore x = 12 12 movies shown each week 5 years ago. Back to question Next Question Back to Excellence Menu

Exercise E1 Question 2 2. A building floor plan has 2 rectangular regions. The left region has one wall completely in common with the right region. The left region has length 5 m longer than its width. The left region has the shared wall ¾ of the width of the right region and the left region is ½ of the length of the right region. If the total perimeter around the building is 186 m, give the dimensions of each of the rectangular regions. Answer Previous Next Question Back to Excellence Menu

Exercise E1 Answer 2 2. Give the dimensions of each of the rectangular regions. Let the width of the left region = x The left region has length 5m longer than its width. Length of left region = x + 5 The left region has the shared wall ¾ of the width of the right region Width of right region = 4x  3 and it is ½ of the length of the right region. Length of right region = 2(x + 5) The total perimeter around the building is 186 m 6(x + 5) + 2(4x  3) = 186 26x  = 186 x = 156 x 3  26 = 18 Therefore the smaller region is 18 m by 23 m and the larger region is 46 m by 24 m. Back to question Insert figure Next Question Back to Excellence Menu

Exercise E1 Question 3 3. A fencing firm have a standard fee that they charge for labour and for the number of metres of fencing they build. They charged one customer \$ for a 165 m long fence that takes 3 hours to erect. Another customer is charged \$ for a fence 42 m long that takes 1.3 hours to erect. What will they estimate as a charge to erect a fence 50 m long that they expect to take 1.5 hours to erect? Answer Previous Next Question Back to Excellence Menu

Exercise E1 Answer 3 3. Fencers have a standard fee that they charge for labour and for the number of metres of fencing they build Let charge rate for labour = \$x per hour and cost per metre of fence = y. They charge one customer \$ for a 165 m long fence that takes 3 hours to erect. 3x + 165y = Another customer is charged \$ for a fence 42 m long that takes 1.3 hours to erect. 1.3x + 42y = To solve these, multiply equation 1 by 1.3: 3.9x y = Multiply equation 2 by -3: -3.9x – 126y = Adding gives: 88.5y = y = Substitute into equation 1 3x = leading to 3x = 85.9 x = 28.3 Therefore cost for labour is \$28.30 per hour and cost per metre of fence is \$14.50. What will they estimate as a charge to erect a fence 50 m long that they expect to take 1.5 hours to erect? Charge = (1.5 x \$28.30) + (50 x \$14.50) = \$767.45 Back to question Next Question Back to Excellence Menu

Exercise E1 Question 4 4. Concrete is needed for a path that is to surround four sides of a rectangular garden. It is to be 0.85 m wide and 0.05 m thick. The garden is 2 metres longer than it is wide. If 0.8 cubic metres of concrete is needed, what is the maximum length the garden can be? Answer Previous Next Question Back to Excellence Menu

Exercise E1 Answer 4 4. What is the length of the garden? Let the length of the garden = x The garden is 2 metres longer than it is wide.. Width of garden = x – 2 Concrete is to be 0.85 m wide and 0.05 m thick. Area of concrete = 2 x 0.85 x x + 2 x 0.85 x (x – 2) + 4 x 0.852 = 1.7x + 1.7x – = 3.4x – 0.51 Volume of concrete = (3.4x – 0.51) x 0.05 = 0.17x – 0.8 cubic metres of concrete is needed 0.17x – < 0.8 0.17x < x < Therefore maximum length is 4.85 m Back to question Figure Ch 01.21 Next Question Back to Excellence Menu

Exercise E1 Question 5 5. A jeweller displays watches in a cabinet. The watches are all the same price. When the cabinet is ¾ full, the value of the cabinet and watches is \$ When the cabinet is 1/3 full, the combined value is \$ What is the cabinet and watches worth when it is full? Answer Previous Next Question Back to Excellence Menu

Exercise E1 Answer 5 5. What is the cabinet and watches worth when it is full? Let value of cabinet = \$c and value of watches = \$w. When the cabinet is ¾ full, the value of the cabinet and watches is valued at \$ c w = When the cabinet is 1/3 full, the combined value is \$ c + w = To solve these, multiply equation 2 by -1 -c – w = Adding gives: w = w = Substitute into equation 1 c = c = 1200 Therefore cost for cabinet full of watches is \$ \$ = \$ Back to question Next Question Back to Excellence Menu

Exercise E1 Question 6 6. A square field has a right-angled isosceles triangle area cut from one corner. The sides of the rectangular field where the corner has been cut off are reduced to 2/3 of their original length. If the resulting field has a perimeter of 96 m, what was the original side lengths of the field? Answer Previous Next Question Back to Excellence Menu

Exercise E1 Answer 6 Back to question 6. What was the original side lengths of the field? Let the length of the original sides = x A square field has a right-angled isosceles triangle area cut from one corner The sides off the rectangular field where the corner has been cut off are reduced to 2/3 of their original length. The resulting field has a perimeter of 96 m Diagonal length = = = x Perimeter = 2x + 2(2/3x) x = x 3.8047x = 96 x = m Original side = m Insert figure Ch 01.22 Next Question Back to Excellence Menu

Exercise E1 Question 7 7. A right-angled triangle section has the longest side 4 m longer than the shortest side. The other side is 2 m longer than the shortest side. What is the area of the section? Answer Previous Next Question Back to Excellence Menu

Exercise E1 Answer 7 7. A right-angled triangle section has the longest side 4 m longer than the shortest side. Let the shortest side = x Longest side = x + 4 The other side is 2 m longer than the shortest side. Length of other side = x + 2 What is the area of the section? Using Pythagoras’ theorem, (x + 4)2 = (x + 2)2 + x2 x2 + 8x = x2 + 4x x2 x2 + 8x = 2x2 + 4x + 4 x2 – 4x – 12 = 0 (x – 6)(x + 2) = 0 x = 6 (since x = -2 not sensible) Area = 0.5  6  8 = 24 m2 Back to question Insert figure Ch 01.23 Next Question Back to Excellence Menu

Exercise E1 Question 8 8. A pattern of tiles is shown in the diagram. If 754 tiles are used, how many layers are there? Answer Previous Next Question Back to Excellence Menu

Exercise E1 Answer 8 8. Number of tiles 4, 10, 18, 28, 40, … Difference between terms is 6, 8, 10, 12, … This is increasing at a constant rate of 2 Therefore quadratic pattern involving 1n2 Creating table shows difference between 1n2 and number of tiles is 3, 6, 9, 12, … Therefore pattern is n2 + 3n If 754 tiles are used, how many layers are there? n2 + 3n = 754 n2 + 3n – 754 = 0 Factors of -754 include -26, 29 (n – 26)(n + 29) = 0 -29 not sensible so 26 layers. Back to question Layers, n Tiles n2 diff 1 4 3 2 10 6 18 9 28 16 12 5 40 25 15 Next Question Back to Excellence Menu

Exercise E1 Question 9 9. If Susan was paid \$30 more a week, she would earn half as much as Sophie. Together they are paid \$510. How much do they each earn? Let m = money that Susan earns. Sophie earns 2  (m + 30) = 2m + 60 Together they earn m + (2m + 60) = 510 3m = 510 3m = (– 60 from both sides) m = ( 3 on both sides) Susan earns \$150 and Sophie earns \$360 Previous Next Question Back to Excellence Menu