If \(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) \(\frac{9}{7(x-3)}\), find the value of R
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Correct Answer: Option C
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
\(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) + \(\frac{9}{7(x-3)}\)
15-2x = R(x - 3) +9(x +4)/7
Put x =-4, we have 15 -2(-4) = -7R
23 -7R;
R = 23/7