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 \(\frac{32}{7}\)
waec 2021furthermathematics

The correct answer is C. \(\frac{23}{7}\)

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

 15-2x = R(x - 3) +9(x +4)/7

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

23 -7R;

R = 23/7

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