2 Solving a linear equation

To solve a linear equation we make the unknown quantity the subject of the equation. We obtain the unknown quantity on its own on the left-hand side. To do this we may apply the same rules used for transposing formulae given in Workbook 1 Section 1.7. These are given again here.

Key Point 2

Operations which can be used in the process of solving a linear equation

∙ add the same quantity to both sides

∙ subtract the same quantity from both sides

∙ multiply both sides by the same quantity

∙ divide both sides by the same quantity

∙ take the reciprocal of both sides (invert)

∙ take functions of both sides; for example cube both sides.

A useful summary of the rules in Key Point 2 is ‘whatever we do to one side of an equation we must also do to the other’.

Example 2

Solve the equation x + 14 = 5 .

Solution

Note that by subtracting 14 from both sides, we leave x on its own on the left. Thus

x + 14 − 14 = 5 − 14 x = − 9

Hence the solution of the equation is x = − 9 . It is easy to check that this solution is correct by substituting x = − 9 into the original equation and checking that both sides are indeed the same. You should get into the habit of doing this.

Example 3

Solve the equation 19 y = 38 .

Solution

In order to make y the subject of the equation we can divide both sides by 19:

19 y = 38 19 y 19 = 38 19 cancelling 19’s gives y = 38 19  so  y = 2

Hence the solution of the equation is y = 2 .

Example 4

Solve the equation 4 x + 12 = 0 .

Solution

Starting from 4 x + 12 = 0 we can subtract 12 from both sides to obtain

4 x + 12 − 12 = 0 − 12  so that  4 x = − 12

If we now divide both sides by 4 we find

4 x 4 = − 12 4 cancelling 4’s gives x = − 3

So the solution is x = − 3 .

Task!

Solve the linear equation 14 t − 56 = 0 .

t = 4

Example 5

Solve the following equations:

  1. x + 3 = 7 ,
  2. x + 3 = − 7 .
Solution
  1. Subtracting 3 from both sides gives x = 7 − 3.
  2. Subtracting 3 from both sides gives x = − 7 − 3 .

Note that when asked to solve x + 3 = ± 7 we can write the two solutions as x = − 3 ± 7 . It is usually acceptable to leave the solutions in this form (i.e. with the 7 term) rather than calculate decimal approximations. This form is known as the surd form .

Example 6

Solve the equation 2 3 ( t + 7 ) = 5 .

Solution

There are a number of ways in which the solution can be obtained. The idea is to gradually remove unwanted terms on the left-hand side to leave t on its own. By multiplying both sides by 3 2 we find

3 2 × 2 3 ( t + 7 ) = 3 2 × 5 = 3 2 × 5 1 and after simplifying and cancelling, t + 7 = 15 2

Finally, subtracting 7 from both sides gives

t = 15 2 − 7 = 15 2 − 14 2 = 1 2

So the solution is t = 1 2 .

Example 7

Solve the equation 3 ( p − 2 ) + 2 ( p + 4 ) = 5 .

Solution

At first sight this may not appear to be in the form of a linear equation. Some preliminary work is necessary. Removing the brackets and collecting like terms we find the left-hand side yields 5 p + 2 so the equation is 5 p + 2 = 5 so that p = 3 5 .

Task!

Solve the equation 2 ( x − 5 ) = 3 − ( x + 6 ) .

  1. First remove the brackets on both sides:

    2 x − 10 = 3 − x − 6 . We may write this as 2 x − 10 = − x − 3 .

  2. Rearrange the equation found in 1 so that terms involving x appear only on the left-hand side, and constants on the right. Start by adding 10 to both sides:

    2 x = − x + 7

  3. Now add x to both sides:

    3 x = 7

  4. Finally solve this to find x :

    7 3

Example 8

Solve the equation

6 1 − 2 x = 7 x − 2

Solution

This equation appears in an unfamiliar form but it can be rearranged into the standard form of a linear equation. By multiplying both sides by ( 1 − 2 x ) and ( x − 2 ) we find

( 1 − 2 x ) ( x − 2 ) × 6 1 − 2 x = ( 1 − 2 x ) ( x − 2 ) × 7 x − 2

Considering each side in turn and cancelling common factors:

6 ( x − 2 ) = 7 ( 1 − 2 x )

Removing the brackets and rearranging to find x we have

6 x − 12 = 7 − 14 x Further rearrangement gives: 20 x = 19

The solution is therefore x = 19 20 .

Example 9

Figure 1 shows three branches of an electrical circuit which meet together at x . Point x is known as a node . As shown in Figure 1 the current in each of the branches is denoted by I , I 1 and I 2 . Kirchhoff’s current law states that the current entering any node must equal the current leaving that node. Thus we have the equation I = I 1 + I 2

Figure 1

No alt text was set. Please request alt text from the person who provided you with this resource.

  1. Given I 2 = 10 A and I = 18 A calculate I 1 .
  2. Suppose I = 36 A and it is known that current I 2 is five times as great as I 1 . Find the branch currents.
Solution
  1. Substituting the given values into the equation we find 18 = I 1 + 10 .

    Solving for I 1 we find

    I 1 = 18 − 10 = 8

    Thus I 1 equals 8 A.

  2. From Kirchhoff’s law, I = I 1 + I 2 .

    We are told that I 2 is five times as great as I 1 , and so we can write I 2 = 5 I 1 .

    Since I = 36 we have

    36 = I 1 + 5 I 1

    Solving this linear equation 36 = 6 I 1 gives I 1 = 6 A.

    Finally, since I 2 is five times as great as I 1 , we have I 2 = 5 I 1 = 30 A.

Exercises

In questions 1-24 solve each equation:

1.   7 x = 14 2.   − 3 x = 6 3.   1 2 x = 7 4.   3 x = 1 2
5.   4 t = − 2 6.   2 t = 4 7.   4 t = 2 8.   2 t = − 4
9.   x 6 = 3 10.   x 6 = − 3 11.   7 x + 2 = 9 12.   7 x + 2 = 23
13.   − 7 x + 1 = − 6 14.   − 7 x + 1 = − 13 15.   17 3 t = − 2 16.   3 − x = 2 x + 8
17.   x − 3 = 8 + 3 x 18.   x 4 = 16 19.   x 9 = − 2 20.   − 13 2 x = 14
21.   − 2 y = − 6 22.   − 7 y = 11 23.   − 69 y = − 690 24.   − 8 = − 4 γ .

In questions 25-47 solve each equation:

25.   3 y − 8 = 1 2 y 26.   7 t − 5 = 4 t + 7 27.   3 x + 4 = 4 x + 3
28.   4 − 3 x = 4 x + 3 29.   3 x + 7 = 7 x + 2 30.   3 ( x + 7 ) = 7 ( x + 2 )
31.   2 x − 1 = x − 3 32.   2 ( x + 4 ) = 8 33. − 2 ( x − 3 ) = 6
34.   − 2 ( x − 3 ) = − 6 35.   − 3 ( 3 x − 1 ) = 2
36.   2 − ( 2 t + 1 ) = 4 ( t + 2 ) 37.   5 ( m − 3 ) = 8
38.   5 m − 3 = 5 ( m − 3 ) + 2 m 39.   2 ( y + 1 ) = − 8
40.   17 ( x − 2 ) + 3 ( x − 1 ) = x 41.   1 3 ( x + 3 ) = − 9 42.   3 m = 4
43.   5 m = 2 m + 1 44.   − 3 x + 3 = 18 45.   3 x + 10 = 31
46.   x + 4 = 8 47.   x − 4 = 23
48.  If y = 2 find x if 4 x + 3 y = 9 49.  If y = − 2 find x if 4 x + 5 y = 3
50.  If y = 0 find x if − 4 x + 10 y = − 8 51.  If x = − 3 find y if 2 x + y = 8
52.  If y = 10 find x when 10 x + 55 y = 530 53.  If γ = 2 find β if 54 = γ − 4 β

In questions 54-63 solve each equation:

54.   x − 5 2 − 2 x − 1 3 = 6 55.   x 4 + 3 x 2 − x 6 = 1 56.   x 2 + 4 x 3 = 2 x − 7
57.   5 3 m + 2 = 2 m + 1 58.   2 3 x − 2 = 5 x − 1 59.   x − 3 x + 1 = 4
60.   x + 1 x − 3 = 4 61.   y − 3 y + 3 = 2 3 62.   4 x + 5 6 − 2 x − 1 3 = x
63.   3 2 s − 1 + 1 s + 1 = 0
64.  Solve the linear equation a x + b = 0 to find x
65.  Solve the linear equation 1 a x + b = 1 c x + d ( a ≠ c ) to find x
1.  2 2.   − 2 3.  14 4.   1 ∕ 6 5.   − 1 ∕ 2 6.  2
7.   1 ∕ 2 8.   − 2 9.  18 10.   − 18 11.  1 12.  3
13.  1 14.  2 15.   − 6 ∕ 17 16.   − 5 ∕ 3 17.   − 11 ∕ 2 18.  64
19.   − 18 20.   − 28 ∕ 13 21.   y = 3 22.   − 11 ∕ 7 23.   y = 10 24.  2
25.   16 ∕ 5 26.  4 27.   1 28.  1/7 29.   5 ∕ 4 30.  7/4
31.   − 2 32.  0 33.  0 34.  6 35.   1 ∕ 9 36.   − 7 ∕ 6
37.   23 ∕ 5 38.  6 39.   − 5 40.  37/19 41.   − 30 42.  3/4
43.   − 5 ∕ 3 44.   − 5 45.  7 46.   8 − 4 47.   23 + 4 48.  3/4
49.  13/4 50.  2 51.  14 52.   − 2 53.   − 13 54.   − 49
55.  12/19 56.  42 57.  1 58.  8/13 59.   − 7 ∕ 3 60.  13/3
61.  15 62.  7/6 63.   − 2 ∕ 5 64.   − b ∕ a 65.   ( d − b ) ( a − c )