SOLVING LINEAR
EQUATIONS
Example
1 Solve 2x  3  11
2x  3  11
take 3 from both sides
2x  3  3  11 3
2x  8
2x 8

2
2
x4
divide both sides by 2
Example
2 Solve 3x  2  16
3x  2  16
add 2 to both sides
3x  2  2  16  2
3x  18
3x 18

3
3
x6
divide both sides by 3
Example
3 Solve 2(3x  1)  4x  5
2(3x  1)  4x  5
expand the brackets
6x  2  4x  5
take 4x from both sides
6x  2  4x  4x  5  4x
2x  2  5
add 2 to both sides
2x  2  2  5  2
2x  7
2x 7

2
2
x3
divide both sides by 2
1
2
Example
4 Solve 12(x  1)  2(2x  9)
12(x  1)  2(2x  9)
expand the brackets
12x  12  4x  18
take 4x from both sides
8x  12  18
add 12 to both sides
8x  30
divide both sides by 8
30
8
3
x3
4
x
Example
5 Solve 2(3x  1)  (2x  5)  15
2(3x  1)  (2x  5)  15
6x  2  2x  5  15
expand the brackets
simplify
4x  7  15
take 7 from both sides
4x  8
divide both sides by 4
x2
Example
5 Solve
7x  5  41 3x
7x  5  41 3x
add 3x to both sides
10x  5  41
add 5 to both sides
10x  46
x
46
10
x  4.6
divide both sides by 10
Example
6 Solve
5  2(x  1)  3x  7
5  2(x  1)  3x  7
expand the brackets
5  2x  2  3x  7
simplify
3  2x  3x  7
add 2x to both sides
3  5x  7
add 7 to both sides
10  5x
divide both sides by 5
x2
Example
7 The three angles of the triangle are shown in terms of x.
a Find the value of x.
b Write down the size of each angle in the triangle.
70  3x
5x  20
a 5x  20  70  3x  x  24  180
3x  114  180
3x  66
simplify
take 114 from both sides
divide both sides by 3
x  22
b 5x  20  5  22  20  110  20  130
70  3x  70  3  22  70  66 
x  24  22  24  46
4
Angles are:
130o, 4o and 46o
NOT TO
SCALE
x  24
Example
8 Find the length of a side of this square.
All lengths are in cm.
4(x  1)
2(3x  2)
2(3x  2)  4(x  1)
expand the brackets
6x  4  4x  4
take 4x from both sides
2x  4  4
add 4 to both sides
2x  8
divide both sides by 2
x4
Length of side = 4(x  1)  4  (4  1)  4  5  20 cm
Example
9 Find the length of the rectangle.
All lengths are in cm.
7x  9
5(7  3x)
7x  9  5(7  3x)
expand the brackets
7x  9  35  15x
add 15x to both sides
22x  9  35
add 9 to both sides
22x  44
divide both sides by 2
x2
Length of rectangle = 7x  9  7  2  9  14  9  5 cm
Example
10 Expression A is 7 more than expression B.
Find the value of x.
A
7(x  4)
B
7(x  4)  7  3(x  5)
expand the brackets
7x  28  7  3x  15
simplify
7x  21 3x  15
take 3x from both sides
4x  21 15
take 21 from both sides
4x  36
x  9
divide both sides by 4
3(x  5)
Example
x
11 Solve
 5  10
3
x
 5  10
3
x
 15
3
3
x
 3  15
3
x  45
add 5 to both sides
multiply both sides by 3
Example
x4
12 Solve
3
7
x4
3
7
7
multiply both sides by 7
x4
73
7
x  4  21
x  17
take 4 from both sides
Example
2x  3
13 Solve
6
5
2x  3
6
5
multiply both sides by 5
2x  3  30
add 3 to both sides
2x  33
x  16
divide both sides by 2
1
2
Example
5
14 Solve
37
x
5
37
x
take 3 from both sides
5
4
x
multiply both sides by x
5  4x
divide both sides by 4
5
x
4
x 1
1
4
Example
15 Solve 6 
6
5
x2
5
x2
multiply both sides by (x + 2)
6(x  2)  5
expand the brackets
6x  12  5
take 12 from both sides
6x  7
x
divide both sides by 6
7
6
x  1
1
6
Example
15 Solve
x6 x4

3
5
x6 x4

3
5
multiply both sides by 15
5 (x  6) 3 (x  4)
15 
 15 
3
5
expand the brackets
5x  30  3x  12
take 3x from both sides
2x  30  12
add 30 to both sides
2x  42
divide both sides by 2
x  21
There is a quicker method for doing the last example.
The quick method can be used when there is a ‘single’ fraction on each side of
the ‘=‘ sign.
x6 x4

3
5
‘cross multiply’
5(x  6)  3(x  4)
expand the brackets
5x  30  3x  12
take 3x from both sides
2x  30  12
add 30 to both sides
2x  42
divide both sides by 2
x  21
Example
16 Solve 8  x 2x  2

2
5
8  x 2x  2

2
5
‘cross multiply’
5(8  x)  2(2x  2)
expand the brackets
40  5x  4x  4
add 5x to both sides
40  9x  4
take 4 from both sides
36  9x
divide both sides by 9
x4
Example
17 Solve 3(x  1) 2x  1

4
3
3(x  1) 2x  1

4
3
‘cross multiply’
3  3(x  1)  4(2x  1)
expand the brackets
9x  9  8x  4
take 8x from both sides
x  9  4
add 9 to both sides
x5