Question

$$\frac{ 1 }{ 3 } (x+1)- \frac{ 3 }{ 4 } (x-2)= \frac{ 1 }{ 6 }$$

Answer

x=4

Solution


Simplify  \(\frac{1}{3}(x+1)\)  to  \(\frac{x+1}{3}\).
\[\frac{x+1}{3}-\frac{3}{4}(x-2)=\frac{1}{6}\]
Simplify  \(\frac{3}{4}(x-2)\)  to  \(\frac{3(x-2)}{4}\).
\[\frac{x+1}{3}-\frac{3(x-2)}{4}=\frac{1}{6}\]
Multiply both sides by \(12\) (the LCM of \(3, 4\)).
\[4(x+1)-9(x-2)=2\]
Expand.
\[4x+4-9x+18=2\]
Simplify  \(4x+4-9x+18\)  to  \(-5x+22\).
\[-5x+22=2\]
Subtract \(22\) from both sides.
\[-5x=2-22\]
Simplify  \(2-22\)  to  \(-20\).
\[-5x=-20\]
Divide both sides by \(-5\).
\[x=\frac{-20}{-5}\]
Two negatives make a positive.
\[x=\frac{20}{5}\]
Simplify  \(\frac{20}{5}\)  to  \(4\).
\[x=4\]