Solve: $8 x + 9 \leq 3 x + 2$

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Solution

given, $8 x + 9 \leq 3 x + 2$

$\Rightarrow 8 x + 9 - 3 x - 2 \leq 0$

$\Rightarrow 5 x + 7 \leq 0$

$\Rightarrow 5 x \leq - 7$

$\Rightarrow x \leq \frac{- 7}{5}$

$∴ x \in (- \infty, \frac{- 7}{5}]$

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Similar questions

3 x + 2 y = 4

8 x + 5 y = 9

8 x + 9 = 3 (x - 1) + 7

9^{| 3 x - 1 |} = 3^{8 x - 2}

Solve 8x + 5y = 9, 3x + 2y = 4

\frac{8 x - 2}{3 x} = 2

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