7 frac{1}{2}-left[2 frac{1}{4} divleft{1 frac | vedclass

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Question

(B) Let the missing value be $x$.First,simplify the expression inside the innermost parenthesis: $(1 \frac{1}{2} - \frac{1}{3} - \frac{1}{6}) = (\frac

Vedclass · Accepted Answer

$\frac{3}{4}$

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(B) Let the missing value be $x$.First,simplify the expression inside the innermost parenthesis: $(1 \frac{1}{2} - \frac{1}{3} - \frac{1}{6}) = (\frac{3}{2} - \frac{1}{3} - \frac{1}{6})$.Finding a common denominator $(6)$: $(\frac{9}{6} - \frac{2}{6} - \frac{1}{6}) = \frac{6}{6} = 1$.Now the equation becomes: $\frac{15}{2} - [\frac{9}{4} \div \{\frac{5}{4} - x(1)\}] = 3$.$\frac{15}{2} - 3 = \frac{9/4}{5/4 - x}$.$\frac{9}{2} = \frac{9/4}{(5-4x)/4}$.$\frac{9}{2} = \frac{9}{5-4x}$.By cross-multiplying or comparing numerators,we get $5 - 4x = 2$.$4x = 3$,so $x = \frac{3}{4}$.

$7 \frac{1}{2}-\left[2 \frac{1}{4} \div\left\{1 \frac{1}{4}-x\left(1 \frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right\}\right]=3$. Find the value of $x$.

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