Solve the inequality for real x: frac{3-2x}{5 | vedclass

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(A) Given the inequality: $\frac{3-2x}{5} < \frac{x}{3} + 2$ Multiply the entire inequality by $15$ (the least common multiple of $5$ and $3$) to clea

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$x \in (-\frac{21}{11}, \infty)$

Vedclass · Answer

(A) Given the inequality: $\frac{3-2x}{5} Multiply the entire inequality by $15$ (the least common multiple of $5$ and $3$) to clear the denominators: $15 	imes \frac{3-2x}{5} $3(3-2x) $9 - 6x Subtract $5x$ from both sides: $9 - 11x Subtract $9$ from both sides: $-11x Divide by $-11$ and reverse the inequality sign: $x > -\frac{21}{11}$ Thus,the solution set is $x \in (-\frac{21}{11}, \infty)$.

Solve the inequality for real $x$: $\frac{3-2x}{5} < \frac{x}{3} + 2$

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