1. Solving Equations X + 3 = 10 4y = 12 2x + 3 = 13 y + 212 3x15 2x + 3 = 11 y – 4 = 7 Great Marlow School Mathematics Department

2. Solving an equation is like keeping a scale balanced. x12Here we can see that x = 12 because the scale is balanced. Great Marlow School Mathematics Department

3. x + 37 On this scale x + 3 = 7 The scale is balanced. But if I take the 3 off… x+3 7 7 The scale is not balanced any more. Great Marlow School Mathematics Department

4. x 7 I can balance it again by taking 3 away from the 7. 4 -3 x This should be written like this to show all the steps needed to work out the answer. X + 3 = 7 - 3 = -3 X = 4 Great Marlow School Mathematics Department

5. y - 25 On this scale y – 2 = 5 Great Marlow School Mathematics Department

6. If I add 2 to the -2 it will become 0. Then the “y” will be on its own, but the scale will not be balanced. Y (– 2 + 2) 5 Remember (-2 + 2 = 0.) Great Marlow School Mathematics Department

7. Y – 2 = 5 + 2 = + 2 Y = 7 Y (– 2 + 2) 5 How can I balance the scale? Yes, add 2 to the other side of the scale. y 7 It should be written like this. Great Marlow School Mathematics Department

8. Now you try some. Remember to show all the steps. 1) x + 3 = 8 - 3 = - 3 x = 5 2) x + 7 = 15 - 7 = - 7 x = 8 3) Y + 5 = 9 - 5 = - 5 y = 4 4) z + 5 = 12 - 5 = - 5 z = 7 5) a + 4 = 15 - 4 = - 4 a = 11 6) Y - 6 = 14 + 6 = +6 y = 20 7) d -5 = 8 + 5 = + 5 d = 13 8) x - 6 = 15 + 6 = + 6 x = 21 9) Y + 13 = 24 - 13 = - 13 y = 11 Great Marlow School Mathematics Department

9. Here 3y = 21 Remember that 3y means 3 X y 3y21 I can’t take the 3 away because it has not been added to the “y”. I must divide the 3y into 3 equal pieces, each piece being 1y. Great Marlow School Mathematics Department

10. 21 3y 3 What’s wrong? Correct – it is not balanced. How can I balance the scale? That’s right – divide 21 into three equal pieces. 3 y 7 21 Great Marlow School Mathematics Department

11. It should be written like this. 3y = 21 3 y = 7 Great Marlow School Mathematics Department

12. How do you think we can solve this equation? Y 212 In this equation y divided by 2 equals 12. Great Marlow School Mathematics Department

13. Y 2 12 Correct! Multiply both sides by 2 X 2X 2 X 2X 2 Remember, if you make a change on one side of the equation, you must make the same change on the other side to keep the equation balanced. y 24 Great Marlow School Mathematics Department

14. Now you try some. 1)3 a = 184) 2b = 7 3 322 a = 6 b = 3.5 2) 5x = 30 5) 7y = 63 55 77 x = 6y = 9 3) 6z = 42 6) 4d = 48 6 4 6 4 z = 7d = 12 7) z = 4 5 z x 5 = 4 x 5 5 z = 20 8) y = 3 8 y x 8 = 3 x 8 8 y = 24 Great Marlow School Mathematics Department

15. 2x + 319 If we knew what 2x was equal to the we could work out what 1x was equal to. What do we need to get rid of? Correct – the +3 Great Marlow School Mathematics Department

16. 2x+ 319 2x -3 Take 3 away from both sides. 16 Now 2x = 16 x 8 Divide both sides by 2. 2x + 3 = 19 -3 = -3 2x = 16 2 2 X = 8 Any questions?

17. 5x + 3 = 48 -3 = -3 5x = 45 5 5 X = 9 6x + 9 = 39 -9 = -9 6x = 30 6 6 X = 5 2x - 3 = 19 +3 = +3 2x = 22 2 2 X = 11 7x - 6 = 29 +6 = +6 7x = 35 7 7 X = 5 3x + 5 = 17 -5 = -5 3x = 12 3 3 X = 4 4x -5 = 19 +5 = +5 4x = 24 4 4 X = 6 1)2) 4) 3) 5) 6) Great Marlow School Mathematics Department

18. 5x + 32x + 15 This time there are x’s on both sides of the balance. Any suggestions? We could take the 2x off both sides 153x + 3 Now it is just like the previous equations. 5x + 3 = 2x +15 -2x = -2x 3x + 3 = 15 -3 = -3 3x = 12 33 X = 4 Great Marlow School Mathematics Department

19. 4x + 2 = 2x +16 -2x = -2x 2x + 2 = 16 -2 = -2 2x = 14 22 X = 7 6x + 8 = 3x +23 -3x = -3x 3x + 8 = 23 -8 = -8 3x = 15 33 X = 5 7x - 10 = 5x +4 -5x = -5x 2x - 10 = 4 +10 = +10 2x = 14 22 X = 7 5x + 14 = 7x +3 -5x = -5x 14 = 2x + 3 -3 = -3 11 = 2x 22 5.5 = x or x = 5.5 9x - 5 = 2x +16 -2x = -2x 7x - 5 = 16 +5 = +5 7x = 21 7 7 X = 3 3x + 8 = x +15 -x = -x 2x + 8 = 15 -8 = -8 2x = 7 22 X = 3.5 Great Marlow School Mathematics Department

20. National curriculum reference: A3d Date: June 1995 Paper: 1 ----------------------------------------- The perimeter of this triangle is 31 cm. Work out the value of c. 2c-3+2c+3+4c-5=31 2c+2c+4c-3+3-5=31 8c-5=31 +5=+5 8c = 36 88 c =4.5 Great Marlow School Mathematics Department

21. Date: June 1997 Paper: 1 ----------------------------------------- (a)Solve 3x = 24 (b)Solve 18 + 3y = 6 - y. [4] Date: June 1999 Paper: 1 ----------------------------------------- (a)Solve the equation 2x = 10. x =..............(1 mark) (b)Solve the equation 6y + 1 = 25. y =...............(2 marks) (c)Solve the equation 8p - 3 = 3p + 13. p =..............(2 marks) See the next slide for the answers. Great Marlow School Mathematics Department

22. 3x = 24 33 X = 8 18 + 3y = 6 – y +y = +y 18 + 4y = 6 -18 = -18 4y = -12 44 Y = -3 2x = 10 22 6y + 1 = 25 - 1 = -1 6y = 24 X = 5 Y = 4 6 6 8p – 3 = 3p + 13 -3p = -3p 5p – 3 = 13 +3 = +3 5p = 16 55 p = 3.2 or 3 1/5 Great Marlow School Mathematics Department