How do you solve $(10)/(x^2-2x)=4/x=5/(x-2)$ ?

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Answer 1 · 2017-05-12T17:41:09

First, we must change the denominators to become the same
( common denominator )

The first thing to do is to expand all the denominators, to fully see the components of the fractions.
$10/(x(x-2))=4/x=5/(x-2)$

If we multiply $4/x$ by $(x-2)/(x-2)$ , and $5/(x-2)$ by $x/x$ , then all three fractions will have the common denominator of $x(x-2)$ :

$(10)/(x(x-2))=(4(x-2))/(x(x-2))=(5(x))/((x-2)(x))$

$10/(x^2-2x)=(4x-8)/(x^2-2x)=(5x)/(x^2-2x)$

multiply all fractions by $(x^2-2x)$

$cancel((x^2-2x))xx10/cancel((x^2-2x))=cancel((x^2-2x))xx(4x-8)/cancel((x^2-2x))=cancel((x^2-2x))xx(5x)/cancel((x^2-2x))$

We are left with

$10=(4x-8)=(5x)$

How do you solve (10)/(x^2-2x)=4/x=5/(x-2)(10)/(x^2-2x)=4/x=5/(x-2)?