If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R...
FURTHER MATHEMATICS
WAEC 2021
If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R
- A. \(\frac{-32}{7}\)
- B. \(\frac{-23}{7}\)
- C. \(\frac{23}{7}\)
- D. \(\frac{32}{7}\)
Correct Answer: C. \(\frac{23}{7}\)
Explanation
\(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) + \(\frac{9}{7(x-3)}\)
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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