How do you solve rational equations $(6x + 4 )/(x+4)=(2x+2)/(x-1)$ ?

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Answer 1 · 2016-02-13T17:05:53

Put it on a common denominator.

The LCD (Least Common Denominator) would be (x + 4)(x - 1)

$((6x + 4)(x- 1))/((x + 4)(x - 1)) = ((2x + 2)(x + 4))/((x + 4)(x - 1))$

We can now eliminate the denominators, because both fractions are now equivalent.

$6x^2 + 4x - 6x - 4 = 2x^2 + 2x + 8x + 8$

Since this is a quadratic equation, we must put everything to one side of the equation, so that the other side is 0.

$4x^2 - 12x - 12 = 0$

$4(x^2 - 3x - 3x) = 0$

As you can see, this cannot be factored. I will solve by completing the square, but you could also use the quadratic formula.

$4(x^2 - 3x + n)= 12$

$n = (b / 2)^2$

$n = (-3/2)^2$

$n = 9/4$

$4(x^2 - 3x + 9/4 - 9/4) = 12$

$4(x^2 - 3x + 9/4) - 4(9/4) = 12$

$4(x - 3/2)^2 - 9 = 12$

$(x - 3/2)^2 = 21/4$

$(x - 3/2) = +-sqrt(21/4)$

$x = (+-sqrt(21))/2 + 3/2$

$x =(+- sqrt(21) + 3)/2$

How do you solve rational equations (6x + 4 )/(x+4)=(2x+2)/(x-1)(6x + 4 )/(x+4)=(2x+2)/(x-1)?