How do you solve $((x-1) / (x^2-x-12)) - 4/(x^2-6x+8) = x/(x^2+x-6)$ ?

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Answer 1 · 2016-08-07T14:52:55

Start by factoring the denominators to see what our denominator will have to be to put on equivalent bases.

$(x - 1)/((x - 4)(x + 3)) - 4/((x - 4)(x - 2)) = x/((x + 3)(x - 2))$

Multiplying to put on a common denominator:

$((x - 1)(x - 2))/((x - 4)(x + 3)(x - 2)) - (4(x + 3))/((x - 4)(x - 2)(x + 3)) = (x(x - 4))/((x + 3)(x - 2)(x - 4))$

We can now eliminate the denominators and solve as a regular equation.

$x^2 - x - 2x + 2 - 4x - 12 = x^2 - 4x$

$x^2 - x^2 -3x - 10 = 0$

$-3x = 10$

$x = -10/3$

Checking in the original equation, we find that $x = -10/3$ satisfies. Hence, the solution set is ${-10/3}$ .

Hopefully this helps!

How do you solve ((x-1) / (x^2-x-12)) - 4/(x^2-6x+8) = x/(x^2+x-6)((x-1) / (x^2-x-12)) - 4/(x^2-6x+8) = x/(x^2+x-6)?