## Precalculus: Mathematics for Calculus, 7th Edition

$x = \frac{1-a}{a^2-a-1}$
$Solve$ $the$ $equation$ $for$ $the$ $indicated$ $variable:$ $a^2x + (a-1) = (a+1)x;$ $for$ $x$ Solve for $x$ Subtract $(a+1)x$ from both sides $a^2x + (a-1) - (a+1)x$ = $(a+1)x - (a+1)x$ Simplify $a^2x + (a-1) - (a+1)x = 0$ Subtract $(a-1)$ from both sides $a^2x + (a-1) - (a+1)x - (a-1)$ = $0 - (a-1)$ Simplify $a^2x - (a+1)x = -(a-1)$ Factor out $x$ from $a^2x - (a+1)x$ $x(a^2 - (a+1)) = -(a-1)$ $x(a^2 - a - 1) = 1-a$ Divide both sides by $a^2 -a -1$ $\frac{x(a^2 - a - 1)}{a^2 - a - 1} = \frac{1-a}{a^2-a-1}$ Simplify $x = \frac{1-a}{a^2-a-1}$