How do you solve $\frac{7}{x + 2} = \frac{2}{x - 5}$ $7/(x+2)= 2/(x-5)$ ?

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How do you solve $\frac{7}{x + 2} = \frac{2}{x - 5}$ $7/(x+2)= 2/(x-5)$ ?

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Answer 1 · 2016-05-02T10:43:18

$x = \frac{39}{5} = 7 \frac{4}{5}$ $x=39/5=7 4/5$

Explanation:

Given
$XXX \frac{7}{x + 2} = \frac{2}{x - 5} XXX$ $color(white)("XXX")7/(x+2)=2/(x-5)color(white)("XXX")$ note: this equation implies $x \neq - 2$ $x!=-2$ and $x \neq 5$ $x!=5$

Multiply both sides by $(x + 2) (x - 5)$ $(x+2)(x-5)$ (sometimes this will be called "cross multiplying")
$XXX 7 (x - 5) = 2 (x + 2)$ $color(white)("XXX")7(x-5)=2(x+2)$
simplify:
$XXX 7 x - 35 = 2 x + 4$ $color(white)("XXX")7x-35=2x+4$
subtract $2 x$ $2x$ from both sides and add $35$ $35$ to both sides
$XXX 5 x = 39$ $color(white)("XXX")5x=39$
divide both sides by $5$ $5$
$XXX x = \frac{39}{5}$ $color(white)("XXX")x=39/5$

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How do you solve $\frac{7}{x + 2} = \frac{2}{x - 5}$ $7/(x+2)= 2/(x-5)$ ?

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How do you solve $\frac{7}{x + 2} = \frac{2}{x - 5}$ $7/(x+2)= 2/(x-5)$ ?