Solve the equation \frac{x+1}{x-1}=\frac{-10}{x+3}+\frac{8}{x^2+2x-3}

Step 1
For the addition (subtraction) between two unlike fractions , first we have to convert both denominators of the fractions to same value. This can be done by finding the LCM(Least Common multiple) of the denominators of the fractions. We can multiply and divide both the fractions by suitable constant to produce same value in both denominators.
The factor of a quadratic equation of the form ${\displaystyle a{x}^2+bx+c=0}$ will be two integers such that whose sum is equal to the value of b and whose product is equal to the value of ${\displaystyle a\cdot c}$.
Consider ${\displaystyle \frac{x+1}{x-1}=\frac{-10}{x+3}+\frac{8}{x^2+2x-3}}$ ................(1)
The trinomial ${\displaystyle x^2+2x-3}$ can be written in factor form as,
${\displaystyle x^2+2x-3=(x+3)(x-1).}$
Thus we can write equation (1) as,
${\displaystyle \frac{x+1}{x-1}=\frac{-10}{x+3}+\frac{8}{x^2+2x-3}}$
Now convert the fractions to like fractions in the RHS side of the last equation. The LCM of the denominators is $(x+3)(x-1)$. There for we have to multiply both numerator and denominator of first fraction in RHS side by $(x-1)$.
Thus we get,
${\displaystyle \frac{x+1}{x-1}=\frac{-10}{x+3}+\frac{8}{(x+3)(x-1)}}$
${\displaystyle \frac{x+1}{x-1}=\frac{-10}{x+3}\times \frac{(x-1)}{(x-1)}+\frac{8}{(x+3)(x-1)}}$
${\displaystyle \frac{x+1}{x-1}=\frac{-10(x-1)+8}{(x+3)(x-1)}}$
Now simplify and cross multiply the denominators, we get
${\displaystyle x+1=\frac{-10x+10+8}{x+3}}$
(x+3)(x+1)=-10x+18
${\displaystyle x^2+x+3x+3=-10x+18}$
${\displaystyle x^2+4x+3+10x-18=0}$
${\displaystyle x^2+14x-15=0}$
Now factorize the quadratic equation ${\displaystyle x^2+14x-15=0}$.
The factors are two integers such that whose sum is equal 14 and whose product is equal to -15 .
Therefore the integers are 15,-1 .
Thus we can write quadratic equation in factor form as,
$(x+15)(x-1)=0$
$x+15=0$
$x=-15$
$x-1=0$
$x=1$
Hence we have $x=1,-15$.
At $x=1$ the equation (1) is not valid. { denominator of equation (1) becomes 0}
Hence the solution of the given equation is $x=-15$.