1. Algebraic Expressions Writing expressions

2. Using letter symbols for unknowns In algebra, we use letter symbols to stand for numbers. These letters are called unknowns or variables. Sometimes we can work out the value of the letters and sometimes we can’t. For example, We can write an unknown number with 3 added on to it as n + 3 This is an example of an algebraic expression.

3. Writing an expression Suppose Jon has a packet of biscuits and he doesn’t know how many biscuits it contains. He can call the number of biscuits in the full packet, b. If he opens the packet and eats 4 biscuits, he can write an expression for the number of biscuits remaining in the packet as: b – 4

4. Writing an equation Jon counts the number of biscuits in the packet after he has eaten 4 of them. There are 22. He can write this as an equation: b – 4 = 22 We can work out the value of the letter b. b = 26 That means that there were 26 biscuits in the full packet.

5. Writing expressions When we write expressions in algebra we don’t usually use the multiplication symbol ×. For example, 5 × n or n × 5 is written as 5 n. The number must be written before the letter. When we multiply a letter symbol by 1, we don’t have to write the 1. For example, 1 × n or n × 1 is written as n.

6. Writing expressions When we write expressions in algebra we don’t usually use the division symbol ÷. Instead we use a dividing line as in fraction notation. For example, When we multiply a letter symbol by itself, we use index notation. For example, n ÷ 3 is written as n 3 n × n is written as n 2. n squared

7. Writing expressions Here are some examples of algebraic expressions: n + 7Add 7 to a number n 5 – n Decrease n from a number 5 2n2n Double the number n or 2 × n 6 n Divide 6 by a number n 4 n + 5Add 5 to 4 lots of a number n n3n3 a number n multiplied by itself 3 times or n × n × n 3 × ( n + 4) or 3( n + 4) a number n is increased by 4 and then the Result is times 3.

8. Writing expressions Miss Green is holding n number of cubes in her hand: She takes 3 cubes away. n – 3 She doubles the number of cubes she is holding. 2 × n or 2n2n Write an expression for the number of cubes in her hand if:

9. Translate the following into algebraic expressions: ① 7 more than x ② x is decreased by two. ③ Double x and add 11 ④ One third of a number ⑤ 3 less than double a number. ⑥ Two numbers have sum 8. If one of them is x, then the other is ………. ⑦ Two numbers are in the ratio 1:3. If the larger one is x then the smaller one is …….. ⑧ Three consecutive integers in ascending order are x, …… and …….. ⑨ The sum of two numbers is 7. One of the numbers is a. What is the other number ( in terms of a) ? ⑩ The total value of x, 20 cent coins and (x-2), 50 cent coins

10. Algebraic Expressions Forming and Solving equations

11. Constructing an equation Ben and Lucy have the same number of sweets. Ben started with 3 packets of sweets and ate 11 sweets. Lucy started with 2 packets of sweets and ate 3 sweets. How many sweets are there in a packet? Let’s call the number of sweets in a packet, n. We can solve this problem by writing the equation: 3 n – 11 The number of Ben’s sweets = is the same as the number of Lucy’s sweets. 2 n – 3

12. Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 3 n – 11 = 2 n – 3 Start by writing the equation down. -2 n Subtract 2 n from both sides. n – 11 = –3 Always line up the equals signs. +11 Add 11 to both sides. n = 8 This is the solution. We can check the solution by substituting it back into the original equation: 3 × 8 – 11 =2 × 8 – 3

13. Constructing an equation I’m thinking of a number. When I multiply the number by 4, I get the same answer as adding 9 to the number. What number am I thinking of? Let’s call the unknown number n. We can solve this problem by writing the equation: The number multiplied by 4 is the same as 4n4n = n + 9 the number plus 9.

14. Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 4 n = n + 9 Start by writing the equation down. -n-n -n-n Subtract n from both sides. 3 n = 9 Always line up the equals signs. ÷3 Divide both sides by 3. n = 3 This is the solution. We can check the solution by substituting it back into the original equation: 4 × 3 =3 + 9

15. Constructing an equation Remember, vertically opposite angles are equal. We can solve this problem by writing the equation: Find the value of x. (2 x + 5) o (65 – 2 x ) o 2 x + 5 = 65 – 2 x

16. Solving the equation Let’s solve this equation by transforming both sides of the equation in the same way. 2 x + 5 = 65 – 2 x +2 x Add 2 x to both sides. 4 x + 5 = 65 -5 Subtract 5 from both sides. 4 x = 60 Divide both sides by 4. ÷4 x = 15 Check: 2 × 15 + 5 =65 – 2 × 15 This is the solution.

17. Rectangle problem The area of this rectangle is 27 cm 2. 8 x – 142 x + 1 Calculate the value of x and use it to find the height of the rectangle. Opposite sides of a rectangle are equal. We can use this fact to write an equation in terms of x.

18. Rectangle problem The area of this rectangle is 27 cm 2. 8 x – 142 x + 1 8 x – 14 = 2 x + 1 –2 x 6 x – 14 = 1 +14 6 x = 15 ÷6 x = 2.5 If x = 2.5 we can find the height of the rectangle using substitution: 8 × 2.5 – 14 =20 – 14 =6 cm

19. Rectangle problem The area of this rectangle is 27 cm 2. What is its width? 8 x – 142 x + 1 Let’s call the width of the rectangle y. y If the height of the rectangle is 6 cm and the area is 27 cm 2 then we can find the width by writing the equation: 6 y = 27 ÷6 y = 4.5 The dimensions of the rectangle are 6 cm by 4.5 cm.

20. Select the correct equation Veronica has 58 cents and buys 4 chocolate bars. They both receive the same amount of change. If c is the cost of one chocolate bar, which equation could we use to solve this problem? A: 4 c + 58 = 7 c + 100B: 58 – 4 c = 1 – 7 c D: 4 c – 58 = 7 c – 1C: 58 – 4 c = 100 – 7 c Thomas has $1 and buys 7 chocolate bars.

21. In pairs now do: Ex 11C ( pg 249) – 2,4,5,7 and 9 Homework : Ex 11B ( pg 246) - 1,2 (only a’s and c’s ), 3a Ex 11 C ( pg 249) – 1, 3, 6 and 8 Remember: Unit 2 Test : December 7 th 2014