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

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

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Answer 1 · 2018-06-27T07:53:28

$x = 4$ $x = 4$

Explanation:

Multiply through by $(x - 2) (x - 3)$ $(x-2)(x-3)$

$\Rightarrow \frac{x (x - 2) (x - 3)}{x - 2} - \frac{(x - 2) (x - 3)}{x - 3} = (x - 2) (x - 3)$ $=> (x(x-2)(x-3))/(x-2) -((x-2)(x-3))/(x-3) = (x-2)(x-3)$

$\Rightarrow x (x - 3) - (x - 2) = (x - 2) (x - 3)$ $=> x(x-3) - (x-2) = (x-2)(x-3)$

$\Rightarrow x^{2} - 3 x - x + 2 = x^{2} - 5 x + 6$ $=> x^2 -3x - x + 2 = x^2 -5x + 6$

$\Rightarrow - 4 x + 2 = - 5 x + 6$ $=> -4x + 2 = -5x + 6$

$\Rightarrow x = 4$ $=> x = 4$

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

1 Answer

Explanation:

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