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

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

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Answer 1 · 2016-03-03T17:58:37

$3 (3 x - 1) - 4 (2 x - 1) = 1 (2 x - 1) (3 x - 1) \to 9 x - 3 - 8 x + 4 = 6 x^{2} - 5 x + 1$ $3(3x-1)-4(2x-1)=1(2x-1)(3x-1) ->9x-3-8x+4=6x^2-5x+1$
$0 = 6 x^{2} - 6 x \to 6 x (x - 1) = 0, 6 x = 0 or x - 1 = 0, x = 0 or x = 1$ $0=6x^2-6x ->6x(x-1)=0,6x=0 or x-1 = 0, x=0 or x=1$

Explanation:

Multiply everything on both sides by the common denominator then put everything on one side and solve for x

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

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