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

Question

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

1 Answer

Impact of this question

1210 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-14T04:47:00

$x = - 1$ $x=-1$

Explanation:

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

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

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

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

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

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

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