Let f(x) = 2x + 4 and g(x) = 6x + 7 here g(x) > 0. Solve the inequality \(\frac{f(x)}{g(x)}\) < 1
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Correct Answer: Option C
Explanation:
\(\frac{f(x)}{g(x)}\) < 1
∴ \(\frac{2x +4}{6x + 7}\) < 1
= 2x + 4 < 6x + 7
= 6x + 7 > 2x + 4
= 6x - 2x > 4 - 7
= 4x > -3
∴ x > -\(\frac{3}{4}\)
\(\frac{f(x)}{g(x)}\) < 1
∴ \(\frac{2x +4}{6x + 7}\) < 1
= 2x + 4 < 6x + 7
= 6x + 7 > 2x + 4
= 6x - 2x > 4 - 7
= 4x > -3
∴ x > -\(\frac{3}{4}\)