Question

$$\frac{1}{x}-\frac{1}{x+1}=\frac{1}{x+2}-\frac{1}{x+3};$$

Answer

x=-3/2

Solution


Multiply both sides by the Least Common Denominator: \(x(x+1)(x+2)(x+3)\).
\[(x+1)(x+2)(x+3)-x(x+2)(x+3)=x(x+1)(x+3)-x(x+1)(x+2)\]
Simplify.
\[{x}^{2}+5x+6={x}^{2}+x\]
Cancel \({x}^{2}\) on both sides.
\[5x+6=x\]
Subtract \(5x\) from both sides.
\[6=x-5x\]
Simplify  \(x-5x\)  to  \(-4x\).
\[6=-4x\]
Divide both sides by \(-4\).
\[-\frac{6}{4}=x\]
Simplify  \(\frac{6}{4}\)  to  \(\frac{3}{2}\).
\[-\frac{3}{2}=x\]
Switch sides.
\[x=-\frac{3}{2}\]

Decimal Form: -1.5