How do you solve $\frac{2 x}{4 - x} = \frac{x^{2}}{x - 4}$ $(2x)/(4-x)=x^2/(x-4)$ ?

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How do you solve $\frac{2 x}{4 - x} = \frac{x^{2}}{x - 4}$ $(2x)/(4-x)=x^2/(x-4)$ ?

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Answer 1 · 2016-08-13T17:33:12

$x = 0$ $x=0$
$x = - 2$ $x=-2$

Explanation:

$\frac{2 x}{4 - x} = \frac{x^{2}}{x - 4}$ $(2x)/(4-x)=x^2/(x-4)$
or
$- \frac{2 x}{x - 4} = \frac{x^{2}}{x - 4}$ $-(2x)/(x-4)=x^2/(x-4)$
or
$x^{2} = - 2 x$ $x^2=-2x$
or
$x^{2} + 2 x = 0$ $x^2+2x=0$
or
$x (x + 2) = 0$ $x(x+2)=0$
or
$x = 0$ $x=0$ ------------------Ans $1$ $1$
or
$x + 2 = 0$ $x+2=0$
or
$x = - 2$ $x=-2$ ----------------------------Ans $2$ $2$

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How do you solve $\frac{2 x}{4 - x} = \frac{x^{2}}{x - 4}$ $(2x)/(4-x)=x^2/(x-4)$ ?

1 Answer

Explanation:

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How do you solve (2x)/(4-x)=x^2/(x-4)2x4−x=x2x−4(2x)/(4-x)=x^2/(x-4)?

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Explanation:

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How do you solve $\frac{2 x}{4 - x} = \frac{x^{2}}{x - 4}$ $(2x)/(4-x)=x^2/(x-4)$ ?