$$\left. \begin{array} { l } { ( 1 ) \frac { 1 } { 3 } ( 2 x - 1 ) - \frac { 1 } { 4 } ( 2 x + 1 ) = \frac { 1 } { 12 } ( 2 - x ) } \end{array} \right.$$
$x=3$
$$\frac{1}{3}\left(2x-1\right)-\frac{1}{4}\left(2x+1\right)=\frac{1}{12}\left(2-x\right)$$
$$\frac{1}{3}\times 2x+\frac{1}{3}\left(-1\right)-\frac{1}{4}\left(2x+1\right)=\frac{1}{12}\left(2-x\right)$$
$$\frac{2}{3}x+\frac{1}{3}\left(-1\right)-\frac{1}{4}\left(2x+1\right)=\frac{1}{12}\left(2-x\right)$$
$$\frac{2}{3}x-\frac{1}{3}-\frac{1}{4}\left(2x+1\right)=\frac{1}{12}\left(2-x\right)$$
$$\frac{2}{3}x-\frac{1}{3}-\frac{1}{4}\times 2x-\frac{1}{4}=\frac{1}{12}\left(2-x\right)$$
$$\frac{2}{3}x-\frac{1}{3}+\frac{-2}{4}x-\frac{1}{4}=\frac{1}{12}\left(2-x\right)$$
$$\frac{2}{3}x-\frac{1}{3}-\frac{1}{2}x-\frac{1}{4}=\frac{1}{12}\left(2-x\right)$$
$$\frac{1}{6}x-\frac{1}{3}-\frac{1}{4}=\frac{1}{12}\left(2-x\right)$$
$$\frac{1}{6}x-\frac{4}{12}-\frac{3}{12}=\frac{1}{12}\left(2-x\right)$$
$$\frac{1}{6}x+\frac{-4-3}{12}=\frac{1}{12}\left(2-x\right)$$
$$\frac{1}{6}x-\frac{7}{12}=\frac{1}{12}\left(2-x\right)$$
$$\frac{1}{6}x-\frac{7}{12}=\frac{1}{12}\times 2+\frac{1}{12}\left(-1\right)x$$
$$\frac{1}{6}x-\frac{7}{12}=\frac{2}{12}+\frac{1}{12}\left(-1\right)x$$
$$\frac{1}{6}x-\frac{7}{12}=\frac{1}{6}+\frac{1}{12}\left(-1\right)x$$
$$\frac{1}{6}x-\frac{7}{12}=\frac{1}{6}-\frac{1}{12}x$$
$$\frac{1}{6}x-\frac{7}{12}+\frac{1}{12}x=\frac{1}{6}$$
$$\frac{1}{4}x-\frac{7}{12}=\frac{1}{6}$$
$$\frac{1}{4}x=\frac{1}{6}+\frac{7}{12}$$
$$\frac{1}{4}x=\frac{2}{12}+\frac{7}{12}$$
$$\frac{1}{4}x=\frac{2+7}{12}$$
$$\frac{1}{4}x=\frac{9}{12}$$
$$\frac{1}{4}x=\frac{3}{4}$$
$$x=\frac{3}{4}\times 4$$
$$x=3$$
Show Solution
Hide Solution
