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WAEC Further Mathematics 2021 Paper

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Question : 36

Total: 38

If

\frac{15 - 2x}{(x+4)(x-3)}

\frac{R}{x+4}

\frac{9}{7(x-3)}

, find the value of R

$\frac{-32}{7}$
$\frac{-23}{7}$
$\frac{23}{7}$
$\frac{32}{7}$

Solution:

$\frac{15 - 2x}{(x+4)(x-3)}$ = $\frac{R}{x+4}$ + $\frac{9}{7(x-3)}$

15-2x = R(x - 3) +9(x +4)/7

Put x =-4, we have 15 -2(-4) = -7R

23 -7R;

R = 23/7

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