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

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(A) Given inequality: $\frac{(2x-1)}{3} \geq \frac{(3x-2)}{4} - \frac{(2-x)}{5}$ Taking the $LCM$ of the denominators on the $RHS$: $\frac{(2x-1)}{3}

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$(-\infty, 2]$

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(A) Given inequality: $\frac{(2x-1)}{3} \geq \frac{(3x-2)}{4} - \frac{(2-x)}{5}$ Taking the $LCM$ of the denominators on the $RHS$: $\frac{(2x-1)}{3} \geq \frac{5(3x-2) - 4(2-x)}{20}$ Expanding the terms: $\frac{(2x-1)}{3} \geq \frac{15x - 10 - 8 + 4x}{20}$ Simplifying: $\frac{(2x-1)}{3} \geq \frac{19x - 18}{20}$ Cross-multiplying: $20(2x - 1) \geq 3(19x - 18)$ $40x - 20 \geq 57x - 54$ Rearranging terms: $-20 + 54 \geq 57x - 40x$ $34 \geq 17x$ Dividing by $17$: $2 \geq x$ or $x \leq 2$ Thus,the solution set is $(-\infty, 2]$.

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

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