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WAEC 2020 · Paper 1 · Q8

If 6x+k2x2+7x−15≡4x+5−22x−3\dfrac{6x + k}{2x^2 + 7x - 15} \equiv \dfrac{4}{x + 5} - \dfrac{2}{2x - 3}, find the value of kk.

Worked solution (try it first)
  1. The denominator factorises as (x+5)(2x−3)(x + 5)(2x - 3), so combine the right-hand side over it.
  2. The numerator becomes 4(2x−3)−2(x+5)4(2x - 3) - 2(x + 5), which is 8x−12−2x−108x - 12 - 2x - 10.
  3. That simplifies to 6x−226x - 22, so comparing constant terms, k=−22k = -22, option B.

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