How do you solve $\frac{3 x + 2}{2 x - 1} = \frac{5 x + 6}{x + 4}$ $(3x+2)/(2x-1)=(5x+6)/(x+4)$ ?

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How do you solve $\frac{3 x + 2}{2 x - 1} = \frac{5 x + 6}{x + 4}$ $(3x+2)/(2x-1)=(5x+6)/(x+4)$ ?

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Answer 1 · 2016-09-25T20:37:53

$x = 2$ $x = 2$ or $x = - 1$ $x = -1$

Explanation:

Multiply both sides of the equation by $(2 x - 1) (x + 4)$ $(2x-1)(x+4)$ to get:

$(3 x + 2) (x + 4) = (5 x + 6) (2 x - 1)$ $(3x+2)(x+4) = (5x+6)(2x-1)$

Multiply out both sides to get:

$3 x^{2} + 14 x + 8 = 10 x^{2} + 7 x - 6$ $3x^2+14x+8 = 10x^2+7x-6$

Subtract the left hand side from both sides to get:

$0 = 7 x^{2} - 7 x - 14$ $0 = 7x^2-7x-14$

$0 = 7 (x^{2} - x - 2)$ $color(white)(0) = 7(x^2-x-2)$

$0 = 7 (x - 2) (x + 1)$ $color(white)(0) = 7(x-2)(x+1)$

Hence $x = 2$ $x=2$ or $x = - 1$ $x=-1$

It remains to check that neither of these values causes a denominator to become $0$ $0$ . The only values that do are $x = \frac{1}{2}$ $x = 1/2$ and $x = - 4$ $x = -4$ so we're good.

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How do you solve $\frac{3 x + 2}{2 x - 1} = \frac{5 x + 6}{x + 4}$ $(3x+2)/(2x-1)=(5x+6)/(x+4)$ ?

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Explanation:

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