This document discusses solving linear equations in one unknown. It provides definitions of key terms like linear equation, solution/root of an equation. It explains that a linear equation in one unknown has only one solution/root. Various forms of linear equations are presented along with the steps to solve each type. Examples are worked through to demonstrate the solving process and common errors are identified. The document also contains sample past year PMR exam questions testing the solving of linear equations.

1. Module PMR
CHAPTER 3 : LINEAR EQUATIONS
1.
Solve linear equation = Finding the value of the unknown which
satisfies the equation.
2.
The solution of the equation is also known as the root of the equation.
3.
A linear equation in one unknown has only one root.
4.
To determine whether a given value is a solution of an equation,
substitute the value into the equation. If the sum of the left hand side
(LHS) = sum of right hand side (RHS), then the given value is a solution.
There are 4 different forms of linear equation as follow:
Equation
Solution
x+a=b
x=b-a
x-a=b
x=b+a
ax = b
x=
x
=b
a
b
a
x=axb
Solving Linear Equations in One Unknown involving combined
operations of +, -, x, ÷ .
Steps:
1. Work on the bracket first, if there is any.
2. Group the terms with the unknown on the left hand side of the
equation while the numbers on the right side.
3. Solve the equation using combined operations.
4. Check your solution by substituting the value into the original
equation.
Examples:
1. Given that 2y + 11 = -5, calculate the value of y.
Solutions:
2 y + 11 = −5
2 y = −5 − 11
2 y = −16
−16
y=
2
y = −8
Linear Equations
32

2. Module PMR
k +5
,find the value of k.
3
2. If k − 1 =
3. Find the value of q which
satisfies the equation
3(q − 4) = q + 2
Solution:
Solution:
k +5
3
3(k − 1) = k + 5
3(q − 4) = q + 2
3k − 3 = k + 5
3q − q = 2 + 12
3k − k = 5 + 3
2k = 8
2q = 14
k −1 =
3q − 12 = q + 2
14
2
q=7
q=
8
2
k =4
k=
4. If 7 - (x + 1) = -4x, then x = ?
Solution:
7 − ( x + 1) = −4 x
7 − x − 1 = −4 x
− x + 4 x = −7 + 1
3 x = −6
−6
3
x = −2
x=
Common Errors
No Errors
Correct Steps
m = −9
x+5 = 3
x = 3+5
x+5 = 3
x = 3−5
x=8
2.
m + 2 = −7
m = −7 − 2
m = −5
1.
m + 2 = −7
m = −7 + 2
x = −2
2y = 8
2y = 8
3.
8
2
y=4
y = 8× 2
y = 16
Linear Equations
y=
33

3. Module PMR
1
p =8
4
4.
p = 8×
1
p =8
4
p = 8× 4
1
4
p = 32
p=2
2x − 3 = 7
2x − 3 = 7
2x = 7 − 3
5.
2x = 7 + 3
2 x = 10
10
x=
2
x=5
2x = 4
x = 4× 2
x=8
Example
1.
Exercise
x + 5 = 11
x = 11 − 5
a) x + 2 = −10
x=6
b) x + 4 = 2
c) 4 + x = 7
d) 10 + x = −3
e) −6 = x + 3
f) −3 = 7 + x
Linear Equations
34

4. Module PMR
2.
x−3 = 5
x = 5+3
a)
b)
x=8
x −8 = 6
x − 10 = −3
c) −7 + x = 2
d) −1 + x = −9
e) 3 = x − 7
f) −5 = −9 + x
3.
5= 6− x
x = 6−5
a) 6 = 7 − x
x =1
b) −5 = 4 − x
c) 3 = − x + 2
d) 1 − x = 10
e) − x + 8 = −3
f) −5 − x = −7
Linear Equations
35

5. Module PMR
4.
−2 x = 10
10
x=
−2
x = −5
a) 3 x = 18
b) −5 x = 25
c) −8 = −2x
d) −16 = 4x
e) − x = −10
f) 3 x =
5.
x
=2
5
x = 2×5
x = 10
a)
1
2
x
= −5
2
b) −
x
=2
7
c)
x 1
=
2 3
d)
5x
= 10
3
2
1
e) − x =
5
4
f) −
Linear Equations
36
x
3
=−
6
2

6. Module PMR
6.
2x −1 = 5
a) 3 x + 4 = −5
2x = 5 + 1
2x = 6
6
x=
2
x=3
b) 6 − 4 x = −2
c)
1
x −3 = 4
2
d) 3 +
2
x = −7
5
e) 4 =
x
+1
2
f) 9 = 3 −
7.
3x + 1 = 2 x − 4
3 x − 2 x = −4 − 1
3
x
2
a) 3 x = 4 x − 7
x = −5
b) 4 x + 9 = 3 + 2 x
c) −5 − x = 11 − 3 x
Linear Equations
37

7. Module PMR
d)
x
−2= x+4
3
e) 2 x =
5x − 3
=x
4
f)
8.
3( x − 2) = 2 x + 5
3x − 6 = 2 x + 5
4 + 9x
5
a)
4( x − 3) − 6 = x
3x − 2 x = 5 + 6
x = 11
b) x − 3( x + 1) = 9
c) x + 4 = 3 − 2( x − 5)
d)
1
(2 x − 3) = −5 + 2 x
3
e)
x −5 x
=
3
6
f)
Linear Equations
38
2 x − 5 3x + 4
=
3
2

8. Module PMR
PMR past year questions
2004
1). Solve each of the following equations.
a) k = −14 − k
3
b) f + (6 − 4 f ) = −31
2
[3 marks]
2005
2). Solve each of the following equations.
a) 2n = 3n − 4
b) 2k =
3 − 7k
5
[3 marks]
2006
3). Solve each of the following equations.
a)
12
=3
n
b) 2(k − 1) = k + 3
[3 marks]
Linear Equations
39

9. Module PMR
2007
4). Solve each of the following equations.
a) x + 10 = 4
b)
5x − 4
=x
3
[3 marks]
2008
5). Solve each of the following equations.
a) p + 5 = −11
b) x − 1 =
x+3
2
[3 marks]
Linear Equations
40

10. Module PMR
CHAPTER 3 :LINEAR EQUATIONS
ANSWERS
a) x = -12
b) x = -2
c) x = 3
d) x = -13
e) x = -9
f) x = -10
a) x = 1
b) x = 9
c) x = -1
d) x = -9
e) x = 11
f) x = 2
3.
5.
7.
2.
a) x = -10
b) x = -14
2
c) x =
3
d) x = 6
5
e) x = −
8
f) x = 9
a) x = 7
b) x = -3
c) x = 8
d) x = -9
e) x = 4
f) x = 3
1.
a) x = 14
b) x = 7
c) x = 9
d) x = -8
e) x = 10
f) x = 4
a) x = 6
b) x = -5
c) x = 4
d) x = -4
e) x = 10
1
f) x =
6
a) x = -3
b) x = 2
c) x = 14
d) x = -25
e) x = 6
f) x = -4
6.
4.
8.
a)
b)
c)
d)
e)
x=6
x = -6
x=3
x=3
x = 10
22
f) −
5
PMR past year questions
2004.
2005.
1). a). k = -7
2). a). n = 4
b). f = 8
b). k =
3
17
2006.
2007.
3). a). n = 4
4). a). x = -6
b). k = 5
2008.
5). a). p = -16
Linear Equations
b). x = 2
b). x = 5
41