How do you solve for x: $12/(x+1) - 10/(x-3)=-3$ ?

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Answer 1 · 2016-06-25T08:33:41

$x = 5 or x= -11/3$

In an equation, we can get rid of the denominators by multiplying every term by the LCM of the denominators.

The LCM is this case is $color(red)((x+1)(x-3))$

$(12xxcolor(red)(cancel((x+1))(x-3)))/cancel((x+1)) -(10xx color(red)((x+1)cancel((x-3))))/cancel((x-3))= -3color(red)((x+1)(x-3))$

$12(x-3) - 10(x+1) = -3(x+1)(x-3))$

$12x-36 -10x-10 = -3(x^2-2x-3)$

$2x-46 = -3x^2+6x+9$

$3x^2-4x-55=0 " now factorise"$

$(3x+11)(x-5)=0$

If $3x+11= 0, rArr x= -11/3$

If $x-5 = rArr x = 5$

How do you solve for x: 12/(x+1) - 10/(x-3)=-312/(x+1) - 10/(x-3)=-3?