$ \frac{3}{4} x-6 \leq x-7 $ Multiplying by 4 on both sides, we get $ 3 x-24 \leq 4 x-28 $ Subtracting $3 x$ from both sides, we get $ -24 \leq x-28 $ Adding 28 on both the sides, we get $ \therefore 4 \leq \mathrm{x} \text { i.e., } \mathrm{x} \geq 4 $ i.e., $\mathrm{x}$ takes all real values greater or equal to 4 . $\therefore$ the solution set is $[4, \infty)$
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