Solving Equations

X + 3 = 10

4y = 12

2x + 3 =

13y + 2

12

3x 15 2x + 3 = 11

y – 4 = 7Great Marlow School Mathematics Department

Solving an equation is like keeping a scale balanced.

x 12 Here we can see that x = 12 because the scale is balanced.

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x + 3 7

On this scale

x + 3 = 7

The scale is balanced.

But if I take the 3 off…

x +3

77

The scale is not balanced any more.

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x7

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 = 4Great Marlow School Mathematics Department

y - 2 5

On this scale y – 2 = 5

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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.)

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Y – 2 = 5 + 2 = + 2Y = 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.

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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

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Here 3y = 21

Remember that 3y means 3 X y

3y 21

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.

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213y3

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

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It should be written like this.

3y = 213 3 y = 7

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How do you think we can solve this equation?

Y2 12

In this equation y divided by 2 equals 12.

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Y2 12

Correct! Multiply both sides by 2

X 2 X 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

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Now you try some.

1)3 a = 18 4) 2b = 73 3 2 2

a = 6 b = 3.5

2) 5x = 30 5) 7y = 63 5 5 7 7x = 6 y = 9

3) 6z = 42 6) 4d = 48 6 46 4z = 7 d = 12

7) z = 45

z x 5 = 4 x 55

z = 20

8) y = 38

y x 8 = 3 x 88

y = 24Great Marlow School Mathematics Department

2x + 3 19

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 +3Great Marlow School Mathematics Department

2x+ 3 19 2x -3

Take 3 away from both sides.

16

Now 2x = 16x 8

Divide both sides by 2.

2x + 3 = 19

-3 = -3

2x = 162 2

X = 8

Any questions?

5x + 3 = 48

-3 = -3

5x = 455 5

X = 9

6x + 9 = 39

-9 = -9

6x = 306 6

X = 5

2x - 3 = 19

+3 = +3

2x = 222 2

X = 11

7x - 6 = 29

+6 = +6

7x = 357 7

X = 5

3x + 5 = 17

-5 = -5

3x = 123 3

X = 4

4x -5 = 19

+5 = +5

4x = 244 4

X = 6

1)

2)

4)

3)

5) 6)

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5x + 3

2x + 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 = 1233

X = 4Great Marlow School Mathematics Department

4x + 2 = 2x +16

-2x = -2x

2x + 2 = 16

-2 = -2

2x = 1422

X = 7

6x + 8 = 3x +23

-3x = -3x

3x + 8 = 23

-8 = -8

3x = 1533

X = 5

7x - 10 = 5x +4

-5x = -5x

2x - 10 = 4

+10 = +10

2x = 1422

X = 7

5x + 14 = 7x +3

-5x = -5x

14 = 2x + 3

-3 = -3

11 = 2x22

5.5 = x or x = 5.5

9x - 5 = 2x +16

-2x = -2x

7x - 5 = 16

+5 = +5

7x = 2177

X = 3

3x + 8 = x +15

-x = -x

2x + 8 = 15

-8 = -8

2x = 722

X = 3.5Great Marlow School Mathematics Department

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 = 368 8

c =4.5

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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

3x = 243 3

X = 8

18 + 3y = 6 – y

+y = +y

18 + 4y = 6

-18 = -18

4y = -124 4Y = -3

2x = 102 2

6y + 1 = 25

- 1 = -1

6y = 24

X = 5

Y = 46 6

8p – 3 = 3p + 13

-3p = -3p

5p – 3 = 13

+3 = +3

5p = 165 5p = 3.2 or 3 1/5

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