If |3x - 2| leq frac{1}{2},then x in | vedclass

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(A) Given the inequality $|3x - 2| \leq \frac{1}{2}$.By the property of absolute value,if $|u| \leq a$,then $-a \leq u \leq a$.So,$-\frac{1}{2} \leq 3

Vedclass · Accepted Answer

$[\frac{1}{2}, \frac{5}{6}]$

Vedclass · Answer

(A) Given the inequality $|3x - 2| \leq \frac{1}{2}$.By the property of absolute value,if $|u| \leq a$,then $-a \leq u \leq a$.So,$-\frac{1}{2} \leq 3x - 2 \leq \frac{1}{2}$.Adding $2$ to all parts of the inequality:$-\frac{1}{2} + 2 \leq 3x \leq \frac{1}{2} + 2$.$\frac{3}{2} \leq 3x \leq \frac{5}{2}$.Dividing by $3$:$\frac{1}{2} \leq x \leq \frac{5}{6}$.Thus,$x \in [\frac{1}{2}, \frac{5}{6}]$.

If $|3x - 2| \leq \frac{1}{2}$,then $x \in$

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