Linear inequalities, equation foundations and four original practice questions with step-by-step answers.
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First, revisit linear equations
These algebra foundations help you work with Grade 9 linear inequalities.
A linear equation contains an unknown value, often written as x, with no powers higher than 1. Solving it means finding the value that makes both sides equal.
Keep the equation balanced: whatever you add, subtract, multiply or divide on one side, do on the other side too. Never divide by zero.
Worked example: solve 3x + 5 = 20
An equation is balanced. Do the same operation on both sides.
1
Subtract 5 from both sides
Undo the +5 to leave 3x on its own.
3x + 5 − 5 = 20 − 5
3x = 15
2
Divide both sides by 3
3x means 3 multiplied by x. Division undoes multiplication.
3x ÷ 3 = 15 ÷ 3
x = 5
3
Check in the original equation
3 × 5 + 5 = 15 + 5 = 20 ✓
Both sides are 20, so x = 5 is correct.
Grade 9 topic: linear inequalities
An inequality compares values instead of saying they are equal. The sign > means “greater than”, and < means “less than”. The signs ≥ and ≤ include equality.
Worked example: solve 3x + 5 > 20
Subtract 5 from both sides: 3x > 15.
Divide both sides by positive 3: x > 5.
The solution is every value greater than 5, not just one value. For example, 6 works because 3 × 6 + 5 = 23 > 20. The value 5 does not work because 20 is not greater than 20.
On a number line, x > 5 has an open circle at 5 and extends to the right. An open circle excludes 5; a filled circle would include it.
Your turn: practice questions
0 of 4 answered
Question 1
Find the value of x.
4x − 7 = 13
Question 2
Find the value of x in this equation with brackets.
2(x + 3) = 18
Question 3
In this example, three identical pens and a KSh 20 exercise book cost KSh 80 altogether. What is the price of one pen? Let p be the price of one pen.
3p + 20 = 80
Question 4
Which of these values of x satisfies the inequality?
3x + 5 > 20
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