Solve for $a : 4 a - 9 > 9 a - 24$

a > 3

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a < 3

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a < - 3

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a < 9

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Solution

The correct option is C $a < 3$
Given, $4 a - 9 > 9 a - 24$
$\Rightarrow 9 a - 4 a < 24 - 9$
$\Rightarrow 5 a < 15$
$\Rightarrow a < 3$

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Similar questions

Δ A B C, A = \frac{2 π}{3}, b - c = 3 \sqrt{3} c m

a r (Δ A B C) = \frac{9 \sqrt{3}}{2} c m^{2}

3 s = - 9

2 \times | 3 x + 9 | < 36

\frac{3^{8} \times a^{6}}{9^{3} \times a^{3}}

9 a^{2} - 24 a + 15 = 0

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