Page 109 - Mathematics

Example 5.42 5
x ≤
8
2
Solve for the value of x if 5x +> 1. 5
Therefore, x ≤ .
Solution 8

2
Given 5x +> 1. Example 5.44 Tanzania Institute of Education
Subtract 2 from both sides of the
inequality. That is, Solve the inequality

5x + 2 2 12− >− 5(x + 2) 3x− ≤ 4 2x+ + 3(x − 1).
5x >− 1
Divide by 5 on both sides Solution
5x >− 1 Given
5 5 5(x + 2) 3x− ≤ 4 2x+ + 3(x − 1).

1 Opening brackets and simplifying
x >−
5 gives,
1
Therefore, x >− . 2x + 10 5x≤ + 1
5
Collecting like terms gives,
Example 5.43 2x − 5x ≤−
1 10
− 3x ≤− 9
Find the value of x which satisfy the Divide by 3− on both sides of the
1
3
inequality –4x +≥ . inequality to get,
2
x ≥ 3
Solution Therefore, x ≥ 3
1
3
Given –4x +≥ 2 .
Subtracting 3 from both sides of the Example 5.45
inequality to obtain,
2
1 Indicate the solution of x ≤ on a
3 3
–4x +− ≥ − 3 number line.
2
5 Solution
–4x ≥−
2 Given x ≤ 2, it implies that
Divide by 4− on both sides of the ± x ≤ 2 Mathematics Form One
inequality to get, Thus, either x+ ≤ 2 or − ≤ 2.
x

x ≤ 2 or x ≥− 2

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Mathematics form 1.indd 103 25/10/2024 09:51:37
Mathematics form 1.indd 103