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BECE Mathematics 2023 Paper

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Question : 10

Total: 40

Solve:

3(x-2)-\;\frac{x}{3}+6>0

$x<0$
$x>0$
$x>3$
$x>\frac{1}{8}$

Solution:

Concept:

Solving a linear inequality in one variable by simplifying and isolating

x

.

Explanation:

Start with the given inequality:

3(x-2)-\frac{x}{3}+6>0

.

Expand the bracket:

3x-6-\frac{x}{3}+6>0

.

The terms

-6

and

+6

cancel each other:

3x-\frac{x}{3}>0

.

Write

3x

as

\frac{9x}{3}

to combine the terms:

\frac{9x-x}{3}>0

.

This simplifies to:

\frac{8x}{3}>0

.

Multiply both sides by

3

to remove the fraction:

8x>0

.

Divide both sides by

8

:

x>0

.

Answer:

The solution is

x>0

, which is option B.

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