Question

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

Answer

x=-1/2

Solution


Simplify  \(\frac{\frac{x}{2}+\frac{3}{4}}{6}\)  to  \(\frac{\frac{x}{2}}{6}+\frac{\frac{3}{4}}{6}\).
\[\frac{\frac{x}{2}}{6}+\frac{\frac{3}{4}}{6}=\frac{x}{2}+\frac{1}{3}\]
Simplify  \(\frac{\frac{x}{2}}{6}\)  to  \(\frac{x}{2\times 6}\).
\[\frac{x}{2\times 6}+\frac{\frac{3}{4}}{6}=\frac{x}{2}+\frac{1}{3}\]
Simplify  \(2\times 6\)  to  \(12\).
\[\frac{x}{12}+\frac{\frac{3}{4}}{6}=\frac{x}{2}+\frac{1}{3}\]
Simplify  \(\frac{\frac{3}{4}}{6}\)  to  \(\frac{3}{4\times 6}\).
\[\frac{x}{12}+\frac{3}{4\times 6}=\frac{x}{2}+\frac{1}{3}\]
Simplify  \(4\times 6\)  to  \(24\).
\[\frac{x}{12}+\frac{3}{24}=\frac{x}{2}+\frac{1}{3}\]
Simplify  \(\frac{3}{24}\)  to  \(\frac{1}{8}\).
\[\frac{x}{12}+\frac{1}{8}=\frac{x}{2}+\frac{1}{3}\]
Multiply both sides by \(12\) (the LCM of \(12, 2\)).
\[x+\frac{3}{2}=6x+4\]
Subtract \(x\) from both sides.
\[\frac{3}{2}=6x+4-x\]
Simplify  \(6x+4-x\)  to  \(5x+4\).
\[\frac{3}{2}=5x+4\]
Subtract \(4\) from both sides.
\[\frac{3}{2}-4=5x\]
Simplify  \(\frac{3}{2}-4\)  to  \(-\frac{5}{2}\).
\[-\frac{5}{2}=5x\]
Divide both sides by \(5\).
\[-\frac{\frac{5}{2}}{5}=x\]
Simplify  \(\frac{\frac{5}{2}}{5}\)  to  \(\frac{5}{2\times 5}\).
\[-\frac{5}{2\times 5}=x\]
Simplify  \(2\times 5\)  to  \(10\).
\[-\frac{5}{10}=x\]
Simplify  \(\frac{5}{10}\)  to  \(\frac{1}{2}\).
\[-\frac{1}{2}=x\]
Switch sides.
\[x=-\frac{1}{2}\]

Decimal Form: -0.5