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WAEC Mathematics 2015 Paper

Question : 39

Total: 45

\;\frac{1}{4}

p + 3q = 10 and 2p - q = 7

Solution:

Concept:

Solve two linear equations by clearing fractions and using substitution.

Explanation:

The given equations are:

\frac{1}{4}p + 3q = 10 \quad \text{(1)}

2p - \frac{1}{3}q = 7 \quad \text{(2)}

Multiply equation (2) by 3 to eliminate the fraction:

6p - q = 21

Rearrange to express

q

in terms of

p

q = 6p - 21

Substitute

q = 6p - 21

into equation (1):

\frac{1}{4}p + 3(6p - 21) = 10

Multiply every term by 4:

p + 12(6p - 21) = 40

p + 72p - 252 = 40

73p = 292

p = \frac{292}{73} = 4

Answer:

p = 4

Therefore, the correct option is A.

4

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