How do you solve $10/x - 12/(x-3) + 4 = 0$ ?

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Answer 1 · 2016-02-12T23:42:44

You must put on a common denominator.

The LCD (Least Common Denominator) is $x(x - 3)$

$(10(x - 3))/(x(x - 3)) - (12(x))/(x(x - 3)) + (4(x^2 - 3x))/(x xx x - 3) = 0$

We can now eliminate the denominators:

$10x - 30 - 12x + 4x^2 - 12x = 0$

$4x^2 - 14x - 30 = 0$

Solve by factoring. Two numbers that multiply to $(-30 xx 4) = -120$ and that add to -14 are -20 and 6.

$4x^2 - 20x + 6x - 30 = 0$

$4x(x - 5) + 6(x - 5) = 0$

$(4x + 6)(x - 5) = 0$

$x = -6/4 and 5$

How do you solve 10/x - 12/(x-3) + 4 = 010/x - 12/(x-3) + 4 = 0?