Solve Rational equations (Quadratic)

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Note that $x \neq-1$ and $x \neq5$

**Answer**

$\frac{8}{x+1}-\frac{18}{x-5}+\frac{3}{\left(x+1\right)\left(x-5\right)}=8$

Or, $8\left(x-5\right)-18\left(x+1\right)+3=8\left(x+1\right)\left(x-5\right)$

Or, $8x-40-18x-18+3=8\left(x^{2}-4x-5\right)$

Or, $-10x-55=8x^{2}-32x-40$

Or, $-10x-55-8x^{2}+32x+40=0$Or, $-8x^{2}+22x-15=0$Or, $8x^{2}-22x+15=0$The solutions of a *quadratic equation* $ax^2+bx + c = 0$ are given by$$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$

In this *quadratic equation*, $a=8$, $b=-22$, and $c=15$

Substituting these values of $a$, $b$ and $c$ in the above formula, we get

$x=\frac{22\pm \sqrt{484-480}}{16}$

or, $x = \frac{22\pm \sqrt{4}}{16}$

or, $x = \frac{22\pm 2}{16}$

or, $x = \frac{24}{16}$ and $x = \frac{20}{16}$

or, $x=\frac{3}{2}$ and $x =\frac{5}{4}$

There are two distinct solutions given by $x = \frac{3}{2}$ and $x = \frac{5}{4}$