How do you solve $\frac{3 x}{x + 1} = \frac{12}{x - 1}$ $(3x )/ (x+1) = 12 / (x-1)$ ?

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How do you solve $\frac{3 x}{x + 1} = \frac{12}{x - 1}$ $(3x )/ (x+1) = 12 / (x-1)$ ?

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Answer 1 · 2018-04-25T02:43:44

the answer is $x = \frac{5 + \sqrt{41}}{2}, \frac{5 - \sqrt{41}}{2}$ $x=(5+sqrt(41))/(2), (5-sqrt(41))/(2)$

Explanation:

first you have to cross multiply and you get: $12 x + 12 = 3 x^{2} - 3 x$ $12x+12=3x^2-3x$ and now move all the terms to one side to make it equal to zero. $3 x^{2} - 15 x - 12 = 0$ $3x^2-15x-12=0$ and you can take out $3$ $3$ from all numbers so: $3 (x^{2} - 5 x - 4) = 0$ $3(x^2-5x-4)=0$ since you can't factor the parenthesis equation, you have to use the quadratic formula which is: $\frac{- b \sqrt{b^{2} - 4 a c}}{2 a}$ $(-bsqrt(b^2-4ac))/(2a)$ and put in the respectful numbers in: $- (- 15) \frac{\sqrt{4 (3) (- 12)}}{2 (3)}$ $-(-15)sqrt(4(3)(-12))/(2(3))$ and now solve to get your answer

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How do you solve $\frac{3 x}{x + 1} = \frac{12}{x - 1}$ $(3x )/ (x+1) = 12 / (x-1)$ ?

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How do you solve (3x )/ (x+1) = 12 / (x-1)3xx+1=12x−1(3x )/ (x+1) = 12 / (x-1)?

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How do you solve $\frac{3 x}{x + 1} = \frac{12}{x - 1}$ $(3x )/ (x+1) = 12 / (x-1)$ ?