3(a + 5) What does this mean? ‘add five to a then multiply the whole lot by three’ Or ‘three lots of a added to three lots of 5

3(a + 5) + 5 a a a

3(a + 5) + 5 a a a 3(a + 5) =

3(a + 5) + 5 a a a 3(a + 5) = (3 x a) +

3(a + 5) + 5 a a a 3(a + 5) = (3 x a) + (3 x 5) =

3(a + 5) + 5 a a a 3(a + 5) = (3 x a) + (3 x 5) = 3a + 15

Expanding Brackets 6(2a + 4) + 4 6(2a + 4) = + 4 (6 x 2a) + (6 x 4) = 12a + 24

Example: 5(2z – 3) Each term inside the brackets is multiplied by the number outside the brackets. Watch out for the signs!

Example: 5(2z – 3) (5 x 2z) + 5 x -3

Example: 5(2z – 3) (5 x 2z) + 5 x - 3 = 10z – 15

Example: 2(3p + 4) + 3(4p + 1)

Example: 2(3p + 4) + 3(4p + 1) = (2 x 3p) + (2 x 4)

Example: 2(3p + 4) + 3(4p + 1) = (2 x 3p) + (2 x 4) + (3 x 4p) + (3 x 1)

Example: 2(3p + 4) + 3(4p + 1) = (2 x 3p) + (2 x 4) + (3 x 4p) + (3 x 1) = 6p + 8

Practice 1:Expand the brackets: (a)(i) 7(n – 3) (ii)4(2x – 3) (iii)p(q – 2p) Multiply out: (3) (a) 5(2y – 3) (1) (c) x(2x +y) (2) Lesson 7n x -12 pq – 2p² 10y x² + xy Are you ready for the answers ?

Practice 2:Expand and simplify: (i)4(x + 5) + 3(x – 7) (2) (ii)5(3p + 2) – 2(5p – 3) (2) Lesson 4x x -21 = 7x p p + 6 = 5p +16 Are you ready for the answers ?

By using substitution answer the following questions: (i)Work out the value of 2a + ay when a = 5 and y = –3 (2) (ii)Work out the value of 5t² - 7 when t=4 (iii) Work out the value of 5x + 1 when x = –3 (iv)Work out the value of D when: (4) D = ut + 2kt If u = 5 t = 1.2 k = –2 (3) Lesson Are you ready for the answers ?

TOP How much do you know? Solve the following (i)x + 5 = 16 (ii)3x + 4 = 19 (2) (b)6y + 9 = 45 (1) (c)2x – 5 = -1 (2) (d) 4(x + 3) = 20 (1) (e) 29 = 9x - 7 (1) (Total 7 marks) Lesson x = 11 x = 5 y = 6 x = 2 x = 4 x = 2 Are you ready for the answers ?

Practice 1: Solve : (a)(i)4x + 2 = 18 (ii)8x – 5 = 19 (iii)7 = 3y - 8 Multiply out the brackets first: (a) 2(x + 3) = 16 (1) (c) 3(2x – 3) = 9 (2) Lesson x = 3 y = 5 x = 5 x = 3 x = 4 Are you ready for the answers ?

Practice 2:Solve: (i)2x + 3 = x + 7 (2) (ii)8r + 3 = 5r + 12 (2) (iii)9x – 14 = 4x + 11 (2) (iv)20y – 16 = 18y - 10 (2) Lesson x = 4 3r = 9 r = 3 5x = 25 x = 5 2y = 6 y = 3 Are you ready for the answers ?

Crossing the equals sign When we take a value across the equals sign we change what it was doing to the opposite. So, if it was + 2 on one side, when we take it to the other it is – 2 If we are x 2 on one side, when we take it to the other it is / 2 For example, x + 5 = 13 x = 13 – 5 x = 8

Using inverse operations to solve equations Solve the following equations using inverse operations. 5 x = 45 x = 45 ÷ 5 x = 9 Check: 5 × 9 = – x = 6 17 = 6 + x 17 – 6 = x Check: 17 – 11 = 6 11 = x x = 11 We usually write the letter before the equals sign.

Using inverse operations to solve equations Solve the following equations using inverse operations. x = 3 × 7 x = 21 Check: 3 x – 4 = 14 3 x = x = 18 Check: 3 × 6 – 4 = 14 x = 18 ÷ 3 x = 6 = 3 x

Balancing equations

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

Solving the equation Move the unknowns (letter terms) to one side and the numbers to the other 3 n – 11 = 2 n – 3 Start by writing the equation down. 3 n – 2 n = – This is the solution. We can check the solution by substituting it back into the original equation: 3  8 – 11 =2  8 – 3 3 n - 2 n – 11 = – 3 n = 8