Question

$$\frac{ { x }^{ 2 } +4x }{ 4 } = \frac{ 6+x }{ 8 } + \frac{ 14+x }{ 8 }$$

Answer

x=2,-5

Solution


Factor out the common term \(x\).
\[\frac{x(x+4)}{4}=\frac{6+x}{8}+\frac{14+x}{8}\]
Join the denominators.
\[\frac{x(x+4)}{4}=\frac{6+x+14+x}{8}\]
Simplify  \(6+x+14+x\)  to  \(20+2x\).
\[\frac{x(x+4)}{4}=\frac{20+2x}{8}\]
Factor out the common term \(2\).
\[\frac{x(x+4)}{4}=\frac{2(10+x)}{8}\]
Simplify  \(\frac{2(10+x)}{8}\)  to  \(\frac{10+x}{4}\).
\[\frac{x(x+4)}{4}=\frac{10+x}{4}\]
Multiply both sides by \(4\).
\[x(x+4)=10+x\]
Expand.
\[{x}^{2}+4x=10+x\]
Move all terms to one side.
\[{x}^{2}+4x-10-x=0\]
Simplify  \({x}^{2}+4x-10-x\)  to  \({x}^{2}+3x-10\).
\[{x}^{2}+3x-10=0\]
Factor \({x}^{2}+3x-10\).
\[(x-2)(x+5)=0\]
Solve for \(x\).
\[x=2,-5\]