Methods and Solving Equations Algebra Methods and Solving Equations

LINEAR PATTERNS Rule: s = 3×n - 2 Rule: d = 4×n + 1 3×1= 3 4×1= 4 + 3 - Linear number patterns are sequences of numbers where the difference between terms is always the same (constant) Rule generating a linear pattern is: Difference × n ± a constant e.g. Write a rule (using n) to describe the following number patterns. n Number of Squares (s) Number of Dots (d) 1 5 2 4 9 3 7 13 10 17 21 3×1= 3 4×1= 4 + 3 + 4 3 = 1 - 2 4 = 5 + 1 + 3 + 4 + 3 + 4 3×4 – 2 4×4 + 1 + 3 + 4 Rule: s = 3×n - 2 Rule: d = 4×n + 1 1. Find the difference between terms and if the same multiply by n 2. Substitute to find constant 3. Check if rule works

SIMPLE QUADRATIC PATTERNS - Quadratic number patterns are sequences of numbers where the difference between terms is not the same and the rule contains a squared term - You need to halve the difference of the differences to find the squared term - i.e. If the difference of the differences is a 2, the rule contains 1n2 e.g. Write a rule for the following pattern Rule: T = 1n2 + 3 n Term (T) 1 4 2 7 3 12 19 5 28 + 3 1×12 = 1 + 2 + 5 1 = 4 + 3 + 2 + 7 + 2 42 + 3 = 19 + 9 3. Halve the 2nd difference to find the n2 rule 1. Find the difference between terms 2. If difference is not the same, find the difference of the differences! 4. Substitute to find constant 5. Check if rule works

HARDER QUADRATIC PATTERNS - The squared term of the rule is found by halving the difference of the differences - i.e. If the difference of the differences is a 6, the rule contains 3n2 - If the simple trial and error does not work, try this technique: Rule: T = 2n2 + 2n + 1 e.g. Write a rule for the following pattern 2×1= 2 n Term (T) 1 5 2 13 3 25 4 41 61 Term (T) – 2n2 2 = 3 + 1 5 - 2×12 3 + 8 + 2 + 4 13 - 2×22 5 + 12 + 2 + 4 7 + 16 + 2 + 4 9 + 20 + 2 11 3. Halve the 2nd difference to find the n2 rule 1. Find the difference between terms 4. Subtract the n2 rule from the term 2. If difference is not the same, find the difference of the differences! 5. Find the linear part of the rule 6. Check if the rule works 2×42 + 2×4 + 1 = 41

POWER RULES a) p10 × p2 = p(10 + 2) b) a3 × a2 × a = a(3 + 2 + 1) 1 1. Multiplication - Does x2 × x3 = x × x × x × x × x ? YES - Therefore x2 × x3 = x5 - How do you get 2 3 = 5 ? + - When multiplying index (power) expressions with the same letter, ADD the powers. No number = 1 i.e. p = 1p1 e.g. Simplify a) p10 × p2 = p(10 + 2) b) a3 × a2 × a = a(3 + 2 + 1) 1 = p12 = a6 - Remember to multiply any numbers in front of the variables first e.g. Simplify a) 2x3 × 3x4 = 2 × 3 x(3 + 4) b) 2a2 × 3a × 5a4 1 = 6 x7 = 2 × 3 × 5 a(2 + 1 + 4) = 30 a7

2. Division - Does 6 = 1 ? 6 YES - Therefore x = 1 x - Does x5 = x × x × x × x × x ? x3 x × x × x YES = x × x × 1 × 1 × 1 - Therefore x5 = x2 x3 - How do you get 5 3 = 2 ? - - When dividing index (power) expressions with the same letter, SUBTRACT the powers. e.g. Simplify a) p5 ÷ p = p(5 - 1) b) x7 x4 = x(7 - 4) 1 = p4 = x3 - Remember to divide any numbers in front of the variables first e.g. Simplify a) 12x5 ÷ 6x4 = 12 ÷ 6 x(5 - 4) b) 5a7 15a2 ÷ 5 = 1 5 a(7 - 2) = 2 x = 1 5 a5 or a5 5 If the power remaining is 1, it can be left out of the answer

a) (c4)6 = c(4 × 6) b) (a3)3 = a(3 × 3) = c24 = a9 a) (3d2)3 = 33 3. Powers of powers - Does (x2)3 = x2 × x2 × x2 ? YES - Does x2 × x2 × x2 = x6 ? YES - Therefore (x2)3 = x6 - How do you get 2 3 = 6 ? × - When taking a power of an index expression to a power, MULTIPLY the powers e.g. Simplify a) (c4)6 = c(4 × 6) b) (a3)3 = a(3 × 3) = c24 = a9 - If there is a number in front, it must be raised to the power, not multiplied e.g. Simplify a) (3d2)3 = 33 × d(2 × 3) b) (2a3)4 = 24 × a(3 × 4) = 27 d6 = 16 a12 - If there is more than one term in the brackets, raise all to the power e.g. Simplify b) (4b2c5)2 = 42 b(2 × 2) c(5 × 2) a) (x3y z4)3 = x(3 × 3) y(1 × 3) z(4 × 3) 1 = x9 y3 z12 = 16 b4 c10

LIKE TERMS e.g. x0 = 1 LIKE terms: 2x, 3x, 31x UNLIKE terms: 2x, 3 4. Powers of zero - Any base to the power of zero has a value of 1 e.g. x0 = 1 LIKE TERMS - LIKE terms are those with exactly the same letter, or combination of letters and powers LIKE terms: 2x, 3x, 31x UNLIKE terms: 2x, 3 4ab, 7ab 5x, 6x2 2ab, 2ac e.g. Circle the LIKE terms in the following groups: a) 3a 5b 6a 2c b) 2xy 4x 12xy 3z 4yx While letters should be in order, terms are still LIKE if they are not.

ADDING/SUBTRACTING TERMS - We ALWAYS aim to simplify expressions from expanded to compact form - Only LIKE terms can be added or subtracted - When adding/subtracting just deal with the numbers in front of the letters e.g. Simplify these expanded expressions into compact form: a) a + a + a 1 1 1 = (1 + 1 + 1)a b) 5x + 6x + 2x = (5 + 6 + 2)x = 3a = 13x c) 3p + 7q + 2p + 5q = (3 + 2)p (+ 7 + 5)q d) 4a + 3a2 + 7a + a2 1 = 5p + 12q = (4 + 7)a (+ 3 + 1)a2 = 11a + 4a2 - For expressions involving both addition and subtraction take note of signs e.g. Simplify the following expressions: a) 4x + 2y – 3x = (4 – 3)x + 2y b) 3a – 4b – 6a + 9b = x + 2y = (3 - 6)a (- 4 + 9)b = -3a + 5b c) 3x2 - 9x + 6x2 + 8x - 5 = (3 + 6)x2 (- 9 + 8)x - 5 = 9x2 - x - 5 If the number left in front of a letter is 1, it can be left out d) 4ab2 +2a2b – 5ab2 + 3ab = -ab2 + 2a2b + 3ab

EXPANDING - Does 6 × (3 + 5) = 6 × 3 + 6 × 5 ? YES 6 × 8 = 18 + 30 48 = 48 - The removal of the brackets is known as the distributive law and can also be applied to algebraic expressions - When expanding, simply multiply each term inside the bracket by the term directly in front e.g. Expand a) 6(x + y) = 6 × x + 6 × y b) -4(x – y) = -4 × x - -4 × y = 6x + 6y = -4x + 4y c) -4(x – 6) = -4 × x - -4 × 6 d) 7(3x – 2) = 7 × 3x - 7 × 2 = -4x + 24 = 21x - 14 e) x(2x + 3y) = x × 2x 1 1 + x × 3y f) -3x(2x – 5) = -3x × 2x - -3x × 5 = 2x2 + 3xy = -6x2 + 15x Don’t forget to watch for sign changes!

- If there is more than one set of brackets, expand them all then collect any like terms. e.g. Expand and simplify a) 2(4x + y) + 8(3x – 2y) = 2 × 4x + 2 × y + 8 × 3x - 8 × 2y = 8x + 2y + 24x - 16y = 32x - 14y b) -3(2a – 3b) – 4(5a + b) = -3 × 2a - -3 × 3b - 4 × 5a + -4 × 1b = -6a + 9b - 20a - 4b = -26a + 5b

SUBSTITUTION e.g. If m = 5, calculate m2 – 4m - 3 = 52 – 4×5 - 3 - Involves replacing variables with numbers and calculating the answer - Remember the BEDMAS rules e.g. If m = 5, calculate m2 – 4m - 3 = 52 – 4×5 - 3 = 25 – 4×5 - 3 = 25 – 20 - 3 = 2 - Formulas can also have more than one variable e.g. If x = 4 and y = 6, calculate 3x – 2y = 3×4 - 2×6 = 12 - 12 = 0 e.g. If a = 2, and b = 5, calculate 2b – a 4 = (2 × 5 – 2) 4 = (10 – 2) 4 Because the top needs to be calculated first, brackets are implied = 8 4 = 2

SQUARE ROOTS ALGEBRAIC FRACTIONS a) b) = 10 x4 = 8 x3 a) b × b2 2 5 - Simply halve the power (as √x is the same as x½ ) e.g. Simplify a) b) = 10 x4 = 8 x3 ALGEBRAIC FRACTIONS 1. Multiplying Fractions - Multiply top and bottom terms separately then simplify. e.g. Simplify: a) b × b2 2 5 = b × b2 b) y2 × 4 3 y = y2 × 4 2 × 5 3 × y = b3 10 = 4y2 3y = 4y 3

a) 2a ÷ a2 5 3 = 2a 5 × 3 a2 = 2a × 3 5 × a2 = 6a 5a2 = 6 5a or 6a-1 5 2. Dividing Fractions - Multiply the first fraction by the reciprocal of the second, then simplify Note: b is the reciprocal of 2 2 b e.g. Simplify: a) 2a ÷ a2 5 3 = 2a 5 × 3 a2 = 2a × 3 5 × a2 = 6a 5a2 = 6 5a or 6a-1 5

a) 3x + 3x 10 10 = 3x + 3x 10 b) 6a - b 5 5 = 6a - b 5 = 6x 10 ÷ 2 4. Adding/Subtracting Fractions a) With the same denominator: - Add/subtract the numerators and leave the denominator unchanged. Simplify if possible. e.g. Simplify: a) 3x + 3x 10 10 = 3x + 3x 10 b) 6a - b 5 5 = 6a - b 5 = 6x 10 ÷ 2 = 3x 5 b) With different denominators: - Multiply denominators to find a common term. - Cross multiply to find equivalent numerators. - Add/subtract fractions then simplify. e.g. Simplify: b) 2x – 5x 3 4 = 3×4 4×2x - 3×5x a) a + 2a 2 3 = 2×3 3×a + 2×2a = 8x – 15x 12 = 3a + 4a 6 = -7x 12 = 7a 6

EXPANDING TWO BRACKETS - To expand two brackets, we must multiply each term in one bracket by each in the second Remember integer laws when multiplying e.g. Expand and simplify a) (x + 5)(x + 2) = x2 + 2x + 5x + 10 b) (x - 3)(x + 4) = x2 + 4x - 3x - 12 = x2 + 7x + 10 = x2 + 1x – 12 To simplify, combine like terms c) (x - 1)(x - 3) = x2 - 3x - 1x + 3 c) (2x + 1)(3x - 4) = 6x2 - 8x + 3x - 4 = x2 – 4x + 3 = 6x2 – 5x – 4

PERFECT SQUARES - When both brackets are exactly the same To simplify, combine like terms e.g. Expand and simplify Watch sign change when multiplying a) (x + 8)2 = (x + 8)(x + 8) b) (x - 4)2 = (x – 4)(x – 4) = x2 + 8x + 8x + 64 = x2 - 4x - 4x + 16 = x2 + 16x + 64 = x2 - 8x + 16 Write out brackets twice BEFORE expanding c) (3x - 2)2 = (3x – 2)(3x – 2) = 9x2 - 6x - 6x + 4 = 9x2 – 12 x + 4

DIFFERENCE OF TWO SQUARES - When both brackets are the same except for signs (i.e. – and +) e.g. Expand and simplify a) (x – 3)(x + 3) = x2 + 3x - 3x - 9 b) (x – 6)(x + 6) = x2 + 6x - 6x - 36 = x2 – 9 = x2 – 36 Like terms cancel each other out c) (2x – 5)(2x + 5) = 4x2 + 10x - 10x - 25 = 4x2 – 25

FACTORISING - Factorising is the reverse of expanding - To factorise: 1) Look for a common factor to put outside the brackets 2) Inside brackets place numbers/letters needed to make up original terms You should always check your answer by expanding it e.g. Factorise a) 2x + 2y = 2( ) x + y b) 2a + 4b – 6c = 2( ) a + 2b - 3c - Always look for the highest common factor e.g. Factorise a) 6x - 15 = 3( ) 2x - 5 b) 30a + 20 = 10( ) 3a + 2 - Sometimes a ‘1’ will need to be left in the brackets e.g. Factorise a) 6x + 3 = 3( ) 2x + 1 b) 20b - 10 = 10( ) 2b - 1

a) 5a2 – 7a5 = a2( ) 5 - 7a3 b) 4b2 + 6b3 = 2b2( ) 2 + 3b - Letters can also be common factors e.g. Factorise a) cd - ce = c( ) d - e b) xyz + 2xy – 3yz = y( ) xz + 2x - 3z c) 4ad – 8a = 4 ( ) a d - 2 - Powers greater than 1 can also be common factors e.g. Factorise a) 5a2 – 7a5 = a2( ) 5 - 7a3 b) 4b2 + 6b3 = 2b2( ) 2 + 3b

Factorising by grouping - When two groups within an expression have their own common factor e.g. Factorise As both 2c and 3d are being multiplied by (a + b) we place them in separate brackets 2ac + 2bc + 3ad + 3bd = 2c( ) a + b + 3d( ) a + b = (2c + 3d)(a + b)

FACTORISING QUADRATICS The general equation for a quadratic is ax2 + bx + c When a = 1 - You need to find two numbers that multiply to give c and add to give b e.g. Factorise To check answer, expand and see if you end up with the original question! a) x2 + 11x + 24 = (x + 3)(x + 8) 1, 24 List pairs of numbers that multiply to give 24 (c) Check which pair adds to give 11 (b) Place numbers into brackets with x 2, 12 3, 8 4, 6 b) x2 + 7x + 6 = (x + 1)(x + 6) 1, 6 List pairs of numbers that multiply to give 6 (c) Check which pair adds to give 7 (b) Place numbers into brackets with x 2, 3

- Expressions can also contain negatives e.g. Factorise a) x2 + x – 12 = (x - 3)(x + 4) b) x2 – 6x – 16 = (x + 2)(x - 8) - 1, 12 1, 16 - As the end number (c) is -12, one of the pair must negative. As the end number (c) is -16, one of the pair must negative. - 2, 6 2, 8 - - 3, 4 4, 4 - Check which pair now adds to give b Make the biggest number of the pair the same sign as b Check which pair now adds to give b Make the biggest number of the pair the same sign as b c) x2 – 9x + 20 = (x - 4)(x - 5) d) x2 – 10x + 25 = (x - 5)(x - 5) - 1, 20 - - 1, 25 - = (x - 5)2 As the end number (c) is +20, but b is – 9, both numbers must be negative - 2, 10 - - 5, 5 - As the end number (c) is +25, but b is – 10, both numbers must be negative - 4, 5 - Check which pair now adds to give b

SPECIAL CASES 1. No end number (c) e.g. Factorise a) x2 + 6x + 0 b) x2 – 10x = x( ) x - 10 0, 6 = x(x + 6) Add in a zero and factorise as per normal OR: factorise by taking out a common factor 2. No x term (b) (difference of two squares) e.g. Factorise a) x2 - 25 + 0x = (x - 5)(x + 5) b) x2 – 100 = (x )(x ) - 10 + 10 -5, 5 c) 9x2 – 121 = (3x )(3x ) - 11 + 11 Add in a zero x term and factorise OR: factorise by using A2 – B2 = (A – B)(A + B)

TWO STAGE FACTORISING When a ≠ 1 1. Common factor - Always try to look for a common factor first. e.g. Factorise a) 2x2 + 12x + 16 = 2( ) x2 + 6x + 8 b) 3x2 – 6x – 9 = 3( ) x2 – 2x – 3 1, 8 = 2(x + 2)(x + 4) 1, 3 - = 3(x + 1)(x – 3) 2, 4 c) 3x2 + 24x = 3x( ) x + 8 d) 4x2 – 36 = 4( ) x2 – 9 = 4(x )(x ) - 3 + 3

2. No common factor (HARD) - Use the following technique e.g. Factorise a) 3x2 – 10x – 8 = 3x2 + 2x – 12x – 8 = x( ) 3x + 2 -4( ) 3x + 2 = (x – 4)(3x + 2) 1x, 24x - Multiply first and last terms Find two terms that multiply to -24x2 but add to -10x Replace 10x with the two new terms Factorise 2 terms at a time. 2x, 12x - 3x, 8x - 4x, 6x - 3x2 × - 8 = -24x2 Write in two brackets b) 2x2 + 7x + 3 = 2x2 + x + 6x + 3 = x( ) 2x + 1 + 3( ) 2x + 1 = (x + 3)(2x + 1) 2x2 × 3 = 6x2 1x, 6x 2x, 3x

RATIONAL EXPRESSIONS - Look to factorise the numerator and/or denominator and then remove the common factor by dividing through. Cancel out common factors e.g. x + 3 = 1 x + 3 e.g. Simplify a) x2 + 7x + 12 x + 3 = (x + 3)(x + 4) x + 3 = (x + 4) As the numerator cancels out, you must leave a ‘1’ on top to signal this b) x – 5 . x2 – 2x – 15 = x – 5 . (x – 5)(x + 3) = 1 . (x + 3)

Don’t forget the integer rules! SOLVING EQUATIONS - Remember that addition/subtraction undo each other as do multiplication/division - Terms containing the variable (x) should be placed on one side (often left) e.g. Solve a) 5x = 3x + 6 b) -6x = -2x + 12 -3x -3x +2x +2x 2x = 6 -4x = 12 Don’t forget the integer rules! ÷2 ÷2 ÷-4 ÷-4 x = 3 x = -3 Always line up equals signs and each line should contain the variable and one equals sign You should always check your answer by substituting into original equation Always look at the sign in front of the term/number to decide operation - Numbers should be placed on the side opposite to the variables (often right) e.g. Solve a) 6x – 5 = 13 b) -3x + 10 = 31 +5 +5 -10 -10 6x = 18 -3x = 21 ÷6 ÷6 ÷-3 ÷-3 x = 3 x = -7

a) 5x + 8 = 2x + 20 b) 4x - 12 = -2x + 24 -2x -2x +2x +2x 3x + 8 = 20 - Same rules apply for combined equations e.g. Solve a) 5x + 8 = 2x + 20 b) 4x - 12 = -2x + 24 -2x -2x +2x +2x 3x + 8 = 20 6x - 12 = 24 -8 -8 +12 +12 3x = 12 6x = 36 ÷3 ÷3 ÷6 ÷6 x = 4 x = 6 - Answers can also be negatives and/or fractions e.g. Solve a) 8x + 3 = -12x - 17 b) 5x + 2 = 3x + 1 +12x +12x -3x -3x 20x + 3 = -17 2x + 2 = 1 -3 -3 -2 -2 20x = -20 2x = -1 ÷20 ÷20 ÷2 ÷2 x = -1 x = -1 2 Make sure you don’t forget to leave the sign too! Answer can be written as a decimal but easiest to leave as a fraction

a) 3(x + 1) = 6 b) 2(3x – 1) = x + 8 3x + 3 = 6 6x - 2 = x + 8 -3 -3 - Expand any brackets first e.g. Solve a) 3(x + 1) = 6 b) 2(3x – 1) = x + 8 3x + 3 = 6 6x - 2 = x + 8 -3 -3 -x -x 3x = 3 5x - 2 = 8 ÷3 ÷3 +2 +2 x = 1 5x = 10 ÷5 ÷5 x = 2 - For fractions, cross multiply, then solve e.g. Solve a) x = 9 4 2 b) 3x - 1 = x + 3 5 2 2x = 36 2(3x - 1) = 5(x + 3) ÷2 ÷2 x = 18 6x - 2 = 5x + 15 -5x -5x x - 2 = 15 +2 +2 x = 17

e.g. Solve 4x - 2x = 10 5 3 ×15 ×15 ×15 60x 5 - 30x 3 = 150 12x - 10x - For two or more fractions, find a common denominator, multiply it by each term, then solve e.g. Solve 4x - 2x = 10 5 3 ×15 ×15 ×15 5 × 3 = 15 60x 5 - 30x 3 = 150 Simplify terms by dividing numerator by denominator 12x - 10x = 150 2x = 150 ÷2 ÷2 x = 75 e.g. Solve 5x - (x + 1) = 2x 6 4 ×24 ×24 ×24 120x 6 - (24x + 24) 4 = 48x 20x - 6x – 6 = 48x 14x – 6 = 48x -48x -48x -34x – 6 = 0 + 6 + 6 -34x = 6 ÷ -34 ÷ -34 x = -6 34

SOLVING INEQUATIONS a) 3x + 8 > 24 b) -2x - 5 ≤ 13 -8 -8 +5 +5 - Inequations contain one of four inequality signs: < > ≤ ≥ - To solve follow the same rules as when solving equations - Except: Reverse the direction of the sign when dividing by a negative e.g. Solve a) 3x + 8 > 24 b) -2x - 5 ≤ 13 -8 -8 +5 +5 3x > 16 -2x ≤ 18 As answer not a whole number, leave as a fraction ÷3 ÷3 ÷-2 ÷-2 Sign reverses as dividing by a negative x > 16 3 x ≥ -9

SOLVING QUADRATICS To solve use the following steps: 1. Move all of the terms to one side, leaving zero on the other 2. Factorise the equation 3. Set each factor to zero and solve. e.g. Solve a) (x + 7)(x – 2) = 0 b) (x – 4)(x – 9) = 0 x + 7 = 0 x – 2 = 0 x – 4 = 0 x – 9 = 0 -7 -7 +2 +2 +4 +4 +9 +9 x = - 7 x = 2 x = 4 x = 9 c) x2 + x – 2 = 0 - 1, 2 d) x2 – 5x + 6 = 0 1, 6 - 2, 3 - (x – 1)(x + 2) = 0 (x + 1)(x – 6) = 0 x – 1 = 0 x + 2 = 0 x + 1 = 0 x – 6 = 0 +1 +1 - 2 - 2 -1 -1 +6 +6 x = 1 x = - 2 x = -1 x = 6

e) x2 + 8x = 0 f) x2 – 11x = 0 x( ) = 0 x + 8 x( ) = 0 x – 11 x = 0 x + 8 = 0 x = 0 x – 11 = 0 - 8 - 8 + 11 + 11 x = -8 x = 11 g) x2 - 49 = 0 h) 9x2 - 4 = 0 (x )(x ) = 0 - 7 + 7 (3x )(3x ) = 0 - 2 + 2 x - 7 = 0 x + 7 = 0 3x - 2 = 0 3x + 2 = 0 +7 +7 - 7 - 7 +2 +2 -2 -2 x = 7 x = -7 3x = 2 3x = -2 ÷3 ÷3 ÷3 ÷3 x = 2/3 x = -2/3 i) x2 = 4x + 5 j) x(x + 3) = 180 -4x -5 -4x -5 1, 5 - x2 + 3x = 180 x2 – 4x – 5 = 0 -180 -180 (x + 1)(x – 5) = 0 x2 + 3x – 180 = 0 (x + 15)(x – 12) = 0 x + 1 = 0 x – 5 = 0 x + 15 = 0 x – 12 = 0 -1 -1 +5 +5 -15 -15 +12 +12 x = -1 x = 5 x = -15 x = 12

WRITING AND SOLVING x + 5 + x + x + 5 + x = 58 4x + 10 = 58 -10 -10 e.g. Write an equation for the following information and solve a) A rectangular pool has a length 5m longer than its width. The perimeter of the pool is 58m. Find its width x + 5 x + 5 + x + x + 5 + x = 58 4x + 10 = 58 Draw a diagram -10 -10 x x 4x = 48 Let x = width ÷4 ÷4 x + 5 x = 12 Therefore width is 12 m b) I think of a number and multiply it by 7. The result is the same as if I multiply this number by 4 and add 15. What is this number? Let n = a number 7 n = n 4 + 15 -4n -4n 3n = 15 ÷3 ÷3 n = 5 Therefore the number is 5

QUADRATIC EQUATIONS - Involves writing an equation from the information then solving e.g. The product of two consecutive numbers is 20. What are they? If x = a number, then the next consecutive number is x + 1 x + 5 x(x + 1) = 20 - 1, 20 x2 + x = 20 - 2, 10 -20 -20 A = 150 m2 x - 4, 5 x2 + x – 20 = 0 (x – 4)(x + 5) = 0 x – 4 = 0 x + 5 = 0 +4 +4 -5 -5 x(x + 5) = 150 x = 4 x = -5 x2 + 5x = 150 The numbers are 4, 5 and -5, -4 -150 -150 x2 + 5x – 150 = 0 (x – 10)(x + 15) = 0 e.g. A paddock of area 150 m2 has a length 5 m longer than its width. Find the dimensions of this paddock. x – 10 = 0 x + 15 = 0 +10 +10 -15 -15 x = 10 x = -15 The dimensions are 10 m by 15 m

REARRANGING FORMULA +2 +2 y + 2 = 6x ÷6 ÷6 y + 2 = x 6 ÷I ÷I R = V I - Involves rearranging the formula in order to isolate the new ‘subject’ - Same rules as for solving are used Remember: When rearranging or changing the subject you are NOT finding a numerical answer e.g. a) Make x the subject of y = 6x - 2 +2 +2 y + 2 = 6x ÷6 ÷6 All terms on the left must be divided by 6 y + 2 = x 6 b) Make R the subject of IR = V Treat letters the same as numbers! ÷I ÷I R = V I c) Make x the subject of y = 2x2 ÷2 ÷2 y = x2 2 Taking the square root undoes squaring

SIMULTANEOUS EQUATIONS - Are pairs of equations with two unknowns To solve we can use one of three methods: 1. ELIMINATION METHOD - Line up equations and either add or subtract so one variable disappears e.g. Solve To remove the ‘y’ variable we add as the signs are opposite. To remove the ‘y’ variable we subtract as the signs are the same. 2x + y = 20 x – y = 4 b) 2x + y = 7 x + y = 4 + ( ) - ( ) 3x = 24 x = 3 ÷3 ÷3 Now we substitute x-value into either equation to find ‘y’ 2 × 3 + y = 7 Now we substitute x-value into either equation to find ‘y’ x = 8 6 + y = 7 2 × 8 + y = 20 -6 -6 16 + y = 20 y = 1 -16 -16 y = 4 Check values in either equation Check values in either equation 8 – 4 = 4 3 + 1 = 4

- You may need to multiply an equation by a number to be able to eliminate a variable e.g. Solve 2x – y = 0 x + 2y = 5 × 2 × 1 4x – 2y = 0 x + 2y = 5 b) 4x – 2y = 28 3x + 3y = 12 × 3 × 4 12x – 6y = 84 12x + 12y = 48 + ( ) - ( ) Multiply the 1st equation by ‘2’ then add to eliminate the y 5x = 5 Multiply the 1st equation by ‘3’ and the 2nd by ‘4’ then subtract to eliminate the x -18y = 36 ÷5 ÷5 ÷-18 ÷-18 x = 1 y = -2 2 × 1 – y = 0 4x – 2 × -2 = 28 2 – y = 0 4x + 4 = 28 Now we substitute x-value into either equation to find ‘y’ Now we substitute y-value into either equation to find ‘x’ -2 -2 -4 -4 – y = -2 4x = 24 ÷-1 ÷-1 ÷4 ÷4 y = 2 x = 6 Check values in either equation Check values in either equation 1 + 2 × 2 = 5 3 × 6 + 3 × -2 = 12 Note that it was possible to eliminate the ‘x’ variable by multiplying second equation by 2 and then subtracting Note that it was possible to eliminate the ‘y’ variable by multiplying the 1st equation by 3 and the 2nd by 2 and then adding

2. SUBSTITUTION METHOD Substitute what the subject equals in for that variable in the other equation - Make x or y the subject of one of the equations - Substitute this equation into the second e.g. Solve a) y = 3x + 1 9x – 2 y = 4 b) x – y = 2 y = 2x + 3 As we are subtracting more than one term, place in brackets and put a one out in front. 9x – 2(3x + 1) = 4 x – (2x + 3) = 2 1 9x – 6x – 2 = 4 x – 2x – 3 = 2 3x – 2 = 4 -x – 3 = 2 +2 +2 +3 +3 3x = 6 -x = 5 ÷3 ÷3 ÷-1 ÷-1 x = 2 x = -5 Now we substitute x-value into first equation to find ‘y’ y = 3×2 + 1 Now we substitute x-value into second equation to find ‘y’ y = 2×-5 + 3 y = 7 y = -7 To check values you can substitute both values into second equation To check values you can substitute both values into second equation