Unit 12 Section 4 : Linear Equations 2

In this section we solve linear equations where more than one step is needed to reach the solution.
Methods of solution vary from one equation to another, but there are two main groups:
  (i) equations where the unknown letter only appears once (like 2x + 7 = 13), and
  (ii) equations where the unknown letter appears more than once (like 5x – 3 = 2x+3).
 

Equations where the unknown letter appears once

Example 1
Look at this equation:
5x + 2 = 17
To solve this, we need to get the x on its own.
Think about what has happened to the x: it has been multiplied by 5 and has had 2 added to it.
To get x on its own, we need to reverse this process, by subtracting 2 and dividing by 5.

Here is the working:
5x + 2=17[subtract 2 from both sides]
5x + 2 – 2=17 – 2[note how the subtract 2 cancels out the add 2]
5x=15[divide both sides by 5]
5x ũ 5=15 ũ 5[note how the divide by 5 cancels out the multiply by 5]
 x=3
Always check your solution works in the original equation: 5 Ũ 3 + 2 = 17

Example 2
Look at this equation:
 p + 3 = 7
2
Think about what has happened to the p this time: it has been divided by 2 and has had 3 added to it.
To get p on its own, we again need to reverse this process, by subtracting 3 and multiplying by 2.

Here is the working:
 p + 3
2
=7[subtract 3 from both sides]
 p + 3 – 3
2
=7 – 3[note how the subtract 3 cancels out the add 3]
 p
2
=4[multiply both sides by 2]
 p Ũ 2
2
=4 Ũ 2[note how the multiply by 2 cancels out the divide by 2]
p=8
Remember to check your solution works in the original equation: 8 ũ 2 + 3 = 7

Example 3
Look at this equation:
5(7 + 2x) = 65
This time, three things have happened to the x: it has been multiplied by 2, it has had 7 added to it, and it
has then been multiplied by 5. To reverse this process, we divide by 5, subtract 7 and then divide by 2.

Here is the working:
5(7 + 2x)=65[divide by 5 on both sides]
5(7 + 2x=13[subtract 7 on both sides]
5(7 + 2x=6[divide by 2 on both sides]
5(7 + 2x=3
Always check your solution works in the original equation: 5 Ũ (7 + 2 Ũ 3) = 65

Note that the last example could also have been solved by multiplying out the brackets first.

 

Equations where the unknown letter appears more than once

Example 4
Look at this equation:
6x – 2 = 4x + 8
To solve this, we need to get the all the "x"s on one side.
There are more "x"s on the left, so we will aim to get all the "x"s on the left and all the "numbers" on the right.

To remove the "4x" on the right, we will subtract 4x on both sides.
To remove the "– 2" on the left, we will add 2 to both sides.

Here is the working:
6x – 2=4x + 8[subtract 4x from both sides] [6x – 4x = 2x]
2x – 2=4x + 8[add 2 to both sides] [8 + 2 = 10]
2x=4x + 10[divide both sides by 2]
2x=4x + 5
As always, check your solution works in the original equation: 6 Ũ 5 – 2 = 4 Ũ 5 + 8

Note that many equations can be solved in more than one way, but all the methods will give the same solution.

 

Practice Question

Work out the answers to the question below then click on the button marked Click on this button below to see the correct answer to see whether you are correct.

Solve these equations:
(a) 4x – 7 = 17     
(b)   p + 3  = 7     
2

 

Exercises

Work out the answers to the questions below and fill in the boxes. Click on the Click this button to see if you are correct button to find out whether you have answered correctly. If you are right then will appear and you should move on to the next question. If appears then your answer is wrong. Click on to clear your original answer and have another go. If you can't work out the right answer then click on Click on this button to see the correct answer to see the answer.

Question 1
Solve the following equations:
(a)3x + 2 = 17
x =
(b)5x – 6 = 9
x =
(c)6x – 4 = 8
x =
(d)3(x + 4) = 30
x =
(e)5(2x – 3) = 15
x =
(f)7 – 2x = 3
x =
(g)6x – 4 = 32
x =
(h)6x + 7 = 1
x =
(i)7x + 6 = 34
x =
(j)6x – 7 = 11
x =
(k)2x + 15 = 16
x =
(l)8 – 2x = 5
x =
(m)38 = 3y + 2
y =
(n)35 = 5(7 + 2p)
p =
(o)56 = 7(2 – 3q)
q =
Question 2
Solve the following equations:
(a)
 x + 5 = 9
2
x =
(b)
14 =  x – 8
3
x =
(c)
 y – 9 = –2
5
y =
(d)
 z + 8 = 3
4
z =
(e)
7 =  p – 6
4
p =
(f)
 x + 5 = 9
2
x =
(g)
14 =  x – 8
3
x =
(h)
 y – 9 = –2
5
y =
(i)
 z + 8 = 3
4
z =
(j)
7 =  p – 6
4
p =
(k)
 2x + 1 = 9
3
x =
(l)
 5x – 7 = 3
4
x =
Question 3
Solve the following equations:
(a)2x + 3 = x + 10
x =
(b)6x – 2 = 4x + 7
x =
(c)16x – 7 = 8x + 17
x =
(d)11x + 2 = 7x + 8
x =
(e)x + 1 = 2x – 2
x =
(f)3x + 12 = 5x – 10
x =
(g)9(x + 7) = 2(5x – 7)
x =
(h)2x + 5 = 7x + 5
x =
Question 4
The formula below can be used to find the perimeter, p, of a rectangle with sides of length x and y.
 p = 2(x + y)

For a particular rectangle, p = 50 and y = 8.

(a) Write the formula out again, replacing p and y with the values given above.

(b) Solve the equation found in part (a) to work out the value of x.
x =

Question 5
A formula states that v = u + at.

(a) Write out the formula, substituting the values v = 10, u = 3 and a = 5.

(b) Solve the equation found in part (a) to work out the value of t.

(c) Write out the formula again, this time substituting the values v = 2, u = 5 and a = 3.

(d) Solve the equation found in part (c) to work out the value of t.

Question 6
The perimeter of the rectangle shown below is 16 cm.
Calculate the value of x.
x = cm
Question 7
The perimeter of the triangle shown below is 23 cm.
Determine the value of x.
x = cm
Question 8
The area of the rectangle shown below is 19.5 cmē.
Determine the value of x.
x =

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