## Algebra 1: Common Core (15th Edition)

The equation has two solutions, $\frac{-2}{3}$ and $3$.
To use the "cross products property" to solve an equation, that equation must be written in the form of two equivalent fractions (one fraction on each side). In this equation: $\frac{1}{x-2}=\frac{2x-6}{x+6}+1$, so we can see that the left side has only one fraction, which satisfies our requirement. But the right side contains more than a single fraction, which makes us unable to use the "cross products property". Therefore, we need to add the sum up so that the right side only contains one fraction. $\frac{2x-6}{x+6}+1=\frac{2x-6}{x+6}+\frac{x+6}{x+6}=\frac{2x-6+x+6}{x+6}=\frac{3x}{x+6}$ Rewriting the equation, we have: $\frac{1}{x-2}=\frac{3x}{x+6}$ Now we can use the "cross products property": $1(x+6)=3x(x-2)$ $x+6=3x^2-6x$ $3x^2-7x-6=0$ $3x^2-9x+2x-6=0$ $3x(x-3)+2(x-3)=0$ $(3x+2)(x-3)=0$ $3x+2=0$ or $x-3=0$ $x=-\frac{2}{3}$ or $x=3$ Check: $\frac{1}{-2/3-2}=\frac{-3}{8}; \frac{2\times-2/3-6}{-2/3+6}+1=-\frac{11}{8}+1=\frac{3}{8}$ $\frac{1}{3-2}=1; \frac{2\times3-6}{3+6}+1=-\frac{0}{9}+1=1$ The equation has two solutions, $\frac{-2}{3}$ and $3$.