# Chapter R - Algebra Reference - R.4 Equations - R.4 Exercises - Page R-16: 34

The solution is $a=1$

#### Work Step by Step

$\dfrac{5}{a}+\dfrac{-7}{a+1}=\dfrac{a^{2}-2a+4}{a^{2}+a}$ Take out common factor $a$ from the denominator of the fraction on the right side of the equation: $\dfrac{5}{a}+\dfrac{-7}{a+1}=\dfrac{a^{2}-2a+4}{a(a+1)}$ Multiply the whole equation by $a(a+1)$: $a(a+1)\Big[\dfrac{5}{a}+\dfrac{-7}{a+1}=\dfrac{a^{2}-2a+4}{a(a+1)}\Big]$ $5(a+1)-7a=a^{2}-2a+4$ $5a+5-7a=a^{2}-2a+4$ Take all terms to the right side and simplify: $0=a^{2}-2a+4-5a-5+7a$ $0=a^{2}-1$ Rearrange: $a^{2}-1=0$ Take $1$ to the right side: $a^{2}=1$ Take the square root of both sides: $\sqrt{a^{2}}=\pm\sqrt{1}$ $a=\pm1$ The original equation is undefined for $a=-1$, so the solution is only $a=1$

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