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WAEC Further Mathematics 2020 Paper

Question : 8

Total: 39

\frac{6x\;+\;k}{2x^{2}\;+\;7x\;-\;15}

\frac{4}{x\;+\;5} - \frac{2}{2x\;-\;3}

. Find the value of k.

Solution:

Concept:

Equate the numerators after making the denominators identical.

Explanation:

Factorise the denominator:

2x^{2} + 7x - 15 = (2x - 3)(x + 5)

So the given equation becomes:

\frac{6x + k}{(2x - 3)(x + 5)} = \frac{4}{x + 5} - \frac{2}{2x - 3}

Combine the right-hand side over the common denominator

(2x - 3)(x + 5)

\frac{4(2x - 3) - 2(x + 5)}{(2x - 3)(x + 5)}

Now equate the numerators:

6x + k = 4(2x - 3) - 2(x + 5)

6x + k = 8x - 12 - 2x - 10

6x + k = 6x - 22

Cancelling

6x

from both sides:

k = -22

Answer:

k = -22

Option B.

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