Question

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

Answer 1

\[{x+2 \over x-3}-{1 \over x}={3 \over x(x-3)} \implies {x(x+2)-(x-3) \over x(x-3)}={3 \over x(x-3)}\]
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 \(x=-1\).

Answer 2

(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