Question

$$\frac{ 1 }{ 3 } (x-1)-(x-3)= \frac{ 1 }{ 3 } (x+3)+ \frac{ 1 }{ 6 }$$

Answer

x=3/2

Solution


Remove parentheses.
\[\frac{1}{3}(x-1)-x+3=\frac{1}{3}(x+3)+\frac{1}{6}\]
Simplify  \(\frac{1}{3}(x-1)\)  to  \(\frac{x-1}{3}\).
\[\frac{x-1}{3}-x+3=\frac{1}{3}(x+3)+\frac{1}{6}\]
Simplify  \(\frac{1}{3}(x+3)\)  to  \(\frac{x+3}{3}\).
\[\frac{x-1}{3}-x+3=\frac{x+3}{3}+\frac{1}{6}\]
Simplify  \(\frac{x+3}{3}\)  to  \(1+\frac{x}{3}\).
\[\frac{x-1}{3}-x+3=1+\frac{x}{3}+\frac{1}{6}\]
Simplify  \(1+\frac{x}{3}+\frac{1}{6}\)  to  \(\frac{x}{3}+\frac{7}{6}\).
\[\frac{x-1}{3}-x+3=\frac{x}{3}+\frac{7}{6}\]
Multiply both sides by \(3\).
\[x-1-3x+9=x+\frac{7}{2}\]
Simplify  \(x-1-3x+9\)  to  \(-2x+8\).
\[-2x+8=x+\frac{7}{2}\]
Add \(2x\) to both sides.
\[8=x+\frac{7}{2}+2x\]
Simplify  \(x+\frac{7}{2}+2x\)  to  \(3x+\frac{7}{2}\).
\[8=3x+\frac{7}{2}\]
Subtract \(\frac{7}{2}\) from both sides.
\[8-\frac{7}{2}=3x\]
Simplify  \(8-\frac{7}{2}\)  to  \(\frac{9}{2}\).
\[\frac{9}{2}=3x\]
Divide both sides by \(3\).
\[\frac{\frac{9}{2}}{3}=x\]
Simplify  \(\frac{\frac{9}{2}}{3}\)  to  \(\frac{9}{2\times 3}\).
\[\frac{9}{2\times 3}=x\]
Simplify  \(2\times 3\)  to  \(6\).
\[\frac{9}{6}=x\]
Simplify  \(\frac{9}{6}\)  to  \(\frac{3}{2}\).
\[\frac{3}{2}=x\]
Switch sides.
\[x=\frac{3}{2}\]

Decimal Form: 1.5