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BECE Mathematics 2013 Paper

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Question : 13

Total: 40

Make

d

the subject of the relation

n=2 d+3

$d=\; \frac{3n}{2}$
$d=\; \frac{n+3}{2}$
$d=\; \frac{n-3}{2}$
$\;\; d=\; \frac{3-n}{2}$

Solution:

Concept:

Rearrange the equation to isolate

d

on one side using inverse operations.

Explanation:

Start with the given relation:

n=2d+3

.

Subtract

3

from both sides to remove the constant term:

n-3=2d

.

Divide both sides by

2

to solve for

d

:

\frac{n-3}{2}=d

.

Thus,

d=\frac{n-3}{2}

.

Answer:

Option C:

d=\frac{n-3}{2}

.

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