Given that \(\frac{6x+m}{2x^{2}+7x-15} \equiv \frac{4}{x+5} - \frac{2}{2x-3}\), find the value of m.
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
Correct Answer: Option D
Explanation:
Taking the LCM of the right hand side of the equation, we have
\(\frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{6x+m}{2x^{2}+7x-15}\)
Comparing the numerators, we have
\(4(2x-3) - 2(x+5) = 6x+m\)
\(8x-12-2x-10 = 6x -22 = 6x + m\)
\(\implies m = -22\)
Taking the LCM of the right hand side of the equation, we have
\(\frac{4(2x-3) - 2(x+5)}{(x+5)(2x-3)} = \frac{6x+m}{2x^{2}+7x-15}\)
Comparing the numerators, we have
\(4(2x-3) - 2(x+5) = 6x+m\)
\(8x-12-2x-10 = 6x -22 = 6x + m\)
\(\implies m = -22\)