anonymous 5 years ago show work too: x+2/x-3 - 1/x = 3/x^2-3x fyi answer is -1

1. anonymous

${x+2 \over x-3}-{1 \over x}={3 \over x(x-3)} \implies {x(x+2)-(x-3) \over x(x-3)}={3 \over x(x-3)}$ Both sides have the same denominator, so for for all values other than $$x=0$$ and $$x=3$$ $x(x+2)-(x-3)=3 \implies x^2+2x-x+3=3 \implies x^2+x=0 \implies x(x+1)=0$ So either $$x=0$$ or $$x=-1$$. But $$x=0$$ is rejected, hence the only solution is $$x=-1$$.

2. gw2011

(x+2)/(x-3) - 1/x = 3/(x^2-3x) (x+2)/(x-3) - 1/x = 3/(x)(x-3) (x+2)/(x-3) - 1/x - 3/(x)(x-3) = 0 Now find a common denominator [(x+2)(x) - (x-3) -3]/)x)(x-3) = 0 (x^2+2x-x+3-3)/(x)(x-3) = 0 (x^2+x)/(x)(x-3) = 0 (x)(x+1)/(x)(x-3) = 0 (x+1)/(x-3) = 0 Multiply both sides of the equation by (x-3) and you get: x+1 = 0 x = -1