Question

\(\dfrac{1}{3}\)-(\(\dfrac{1}{2}\)x+\(\dfrac{1}{4}\))=\(\dfrac{7}{6}\)

Answer 1

\(\dfrac{1}{3}-\left(\dfrac{1}{2}x+\dfrac{1}{4}\right)=\dfrac{7}{6}\)

\(\Rightarrow-\dfrac{1}{2}x=\dfrac{14}{12}-\dfrac{4}{12}+\dfrac{3}{12}\)

\(\Rightarrow-\dfrac{1}{2}x=\dfrac{13}{12}\)

\(\Rightarrow x=\dfrac{13\cdot\left(-2\right)}{12}=-\dfrac{26}{12}=-\dfrac{13}{6}\)

Answer 2

\(\dfrac{1}{3}-\left(\dfrac{1}{2}x+\dfrac{1}{4}\right)=\dfrac{7}{6}\)

\(=>\dfrac{1}{2}x+\dfrac{1}{4}=\dfrac{1}{3}-\dfrac{7}{6}\)

\(=>\dfrac{1}{2}x+\dfrac{1}{4}=-\dfrac{5}{6}\)

\(=>\dfrac{1}{2}x=-\dfrac{5}{6}-\dfrac{1}{4}\)

\(=>\dfrac{1}{2}x=-\dfrac{13}{12}\)

\(=>x=\dfrac{13}{12}:\dfrac{1}{2}\)

\(=>x=-\dfrac{13}{6}\)