Page 110 - Mathematics_F1

Example 5.41 5
x ≤
8
Solve for the value of x if 5x +> 1. 5
2
Therefore, x ≤ .
Solution 8
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Given 5x +> 1. Example 5.43 Tanzania Institute of Education
2
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
x >− Opening brackets and simplifying
5 gives,
1
Therefore, x >− . 2x + 10 5x≤ + 1
5
Collecting like terms gives,
Example 5.42
1 10
2x − 5x ≤−
− 3x ≤− 9
Find the value of x which satisfy the
1 Divide by 3− on both sides of the
3
inequality –4x +≥ .
2 inequality to get, x ≥ 3.
Solution Therefore, x ≥ 3.
1
3
Given –4x +≥ 2 . Example 5.44
Subtracting 3 from both sides of the
2
inequality to obtain, Indicate the solution of x ≤ on a
1 number line.
3 3
–4x +− ≥ − 3
2 Solution
5
–4x ≥− Given x ≤ 2, it implies that
2 ± x ≤ 2
Divide by 4− on both sides of the Thus, either x+ ≤ 2 or − ≤ 2. Mathematics Form One
x
inequality to get,

x ≤ 2 or x ≥− 2

103

25/09/2025 15:01:19
Mathematics form 1.indd 103 25/09/2025 15:01:19
Mathematics form 1.indd 103