Question

How can I solve this equation

`(5)/(x+1)+(1)/(x-3)=(-6)/(x^(2)-2x-3 )`

Answer 1

`(5)/(x+1)+(1)/(x-3)=(-6)/(x^2-2x-3)\\\\(5(x-3))/((x+1)(x-3))+(1(x+1))/((x+1)(x-3))=(-6)/(x^2-2x-3)\ \ \ \ |\text{use distributive property}\\\\((5)(x)+(5)(-3)+(1)(x)+(1)(1))/((x)(x)+(x)(-3)+(1)(x)+(1)(-3))=(-6)/(x^2-2x-3)\\\\(5x-15+x+1)/(x^2-3x+x-3)=(-6)/(x^2-2x-3)\\\\((5x+x)+(-15+1))/(x^2-2x-3)=(-6)/(x^2-2x-3)\\\\(6x-14)/(x^2-2x-3)=(-6)/(x^2-2x-3)`

The fractions with the same denominators are equal if nominators are equal.

Therefore we have the equation:

`6x-14=-6\ \ \ \ |\text{add 14 to both sides}\\\\6x=8\ \ \ \ |\text{divide both sides by 6}\\\\x=(8)/(6)\\\\\boxed{x=(4)/(3)}`