Solve the given inequality for real x: frac{1 | vedclass

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(A) Given inequality: $\frac{1}{2}\left(\frac{3x}{5}+4\right) \geq \frac{1}{3}(x-6)$Multiply both sides by $6$ to clear the denominators:$3\left(\frac

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$(-\infty, 120]$

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(A) Given inequality: $\frac{1}{2}\left(\frac{3x}{5}+4\right) \geq \frac{1}{3}(x-6)$Multiply both sides by $6$ to clear the denominators:$3\left(\frac{3x}{5}+4\right) \geq 2(x-6)$Expand the terms:$\frac{9x}{5} + 12 \geq 2x - 12$Rearrange the terms to isolate $x$:$12 + 12 \geq 2x - \frac{9x}{5}$$24 \geq \frac{10x - 9x}{5}$$24 \geq \frac{x}{5}$Multiply by $5$:$120 \geq x$Thus,all real numbers $x$ such that $x \leq 120$ are solutions.The solution set is $(-\infty, 120]$.

Solve the given inequality for real $x$: $\frac{1}{2}\left(\frac{3x}{5}+4\right) \geq \frac{1}{3}(x-6)$

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