Algebraic Substitution: GCSE Revision with Examples
One of the first steps of algebra is understanding that ‘letters’, known in maths as variables or unknowns, can be used to represent unknown numbers, or values.
Being able to substitute values into expressions is an important skill in algebra and will help you gain a better understanding of using formulae and solving equations later in the course.
In this lesson, we will learn how to substitute positive and negative numbers into algebraic expressions. This will include fractions and decimals.
What is Substitution in Algebra?
Substitution in mathematics means to replace the variables in an expression with their number values. After substituting, we can then work out the value of the whole expression.
Substitution is important when using formulae to find real-life values to questions involving area, length, speed, and distance, etc.
Examples of Algebraic Substitution
Example 1:
Find the value of 4a + 3 when a = 5
Solution:
- 4a means 4 × a
- a = 5, so 4a = 4 × 5 = 20
- 4a + 3 = 20 + 3 = 23
Answer: 23
Example 2:
Find the value of 3x + 7 when x = 6
Solution:
- Substitute the value of x: 3x + 7 = 3 × 6 + 7
- Calculate: 3 × 6 + 7 = 18 + 7 = 25
Answer: 25
Example 3:
Find the value of 2x – 10 when x = -7
Solution:
- Substitute the value of x: 2x – 10 = 2(-7) – 10
- Calculate: 2(-7) – 10 = -14 – 10 = -24
Answer: -24
Example 4:
Find the value of -3x + 10y when x = -5 and y = 8
Solution:
- Substitute: -3x + 10y = -3 × -5 + 10 × 8
- Calculate: -3 × -5 + 10 × 8 = 15 + 80 = 95
Answer: 95
Example 5:
Find the value of (x – 1.7)² – y when x = 2.9 and y = 0.8
Solution:
- Substitute: (x – 1.7)² – y = (2.9 – 1.7)² – 0.8
- Calculate: (2.9 – 1.7)² – 0.8 = 1.2² – 0.8 = 1.44 – 0.8 = 0.64
Answer: 0.64
Example 6:
Find the value of 6ab when a = 1/3 and b = 1/4
Solution:
- Substitute: 6ab = 6 × 1/3 × 1/4
- Calculate: 6 × 1/3 × 1/4 = 6 × 1/12 = 6/12 = 1/2
Answer: 1/2
Substitution with Fractions
Find the value of \( 6ab \) when \( a = \frac{1}{3} \) and \( b = \frac{1}{4} \)
Solution:
- Substitute: \( 6ab = 6 \times \frac{1}{3} \times \frac{1}{4} \)
- Calculate: \( 6 \times \frac{1}{3} \times \frac{1}{4} = 6 \times \frac{1}{12} = \frac{6}{12} = \frac{1}{2} \)
Answer: \( \frac{1}{2} \)
Substitution With Three Variables
Find the value of \( 2f(3g - h) \) when \( f = 15 \), \( g = -2 \), and \( h = 4 \)
Solution:
- Substitute: \( 2f(3g - h) = 2 \times 15(3 \times (-2) - 4) \)
- Calculate: \( 2 \times 15(3 \times (-2) - 4) = 2 \times 15(-6 - 4) = 2 \times 15(-10) = 2 \times -150 \)
Answer: \( -300 \)
Substitution With Powers and Roots
Find the value of \( 23(\sqrt{a} + b)c \) when \( a = 9 \), \( b = 2 \), and \( c = 2 \)
Solution:
- Substitute: \( 23(\sqrt{9} + 2)^{2} \)
- Following the rules for order of operations, first find the value of the expression inside the brackets, then square this value and multiple by 23.
- Calculate: \( 23(\sqrt{9} + 2)^{2} = 23(3 + 2)^{2} = 23(5)^{2} = 23(25) = 23 \times 25 \)
Answer: \( 575 \)
Algebraic Substitution Practice Questions
Question 1
Find the value of \( 4a + 3 \) when \( a = 5 \)