How do you solve $\frac { x - 1} { x + 3} = \frac { x + 2} { x }$ ?

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Answer 1 · 2017-06-14T04:47:00

$x=-1$

Since this is set up like a proportion, cross multiply.
$(x-1)/(x+3) = (x+2)/x$
$(x)(x-1)=(x+2)(x+3)$

Multiply out both sides, respectively.
$(x)(x-1)=(x+2)(x+3)$
$x^2-x=x^2+3x+2x+6$
$x^2-x=x^2+5x+6$

Begin isolating $x$ to one side by subtracting both sides by $x^2$ .
$x^2-x=x^2+5x+6$
$-x=5x+6$

Subtract both sides by $5x$ to get $x$ all onto the left side.
$-x=5x+6$
$-6x=6$

Divide by $-6$ on both sides to isolate $x$ , leaving you with the solution of the equation.
$-6x=6$
$x=-1$

How do you solve \frac { x - 1} { x + 3} = \frac { x + 2} { x }\frac { x - 1} { x + 3} = \frac { x + 2} { x }?