Solve Rational equations (Quadratic)

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Note that $x \neq5$ and $x \neq4$

**Answer**

$\frac{44}{x-5}-\frac{25}{x-4}-\frac{4}{\left(x-5\right)\left(x-4\right)}=-3$

Or, $44\left(x-4\right)-25\left(x-5\right)-4=-3\left(x-5\right)\left(x-4\right)$

Or, $44x-176-25x+125-4=-3\left(x^{2}-9x+20\right)$

Or, $19x-55=-3x^{2}+27x-60$

Or, $19x-55+3x^{2}-27x+60=0$Or, $3x^{2}-8x+5=0$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*, $a=3$, $b=-8$, and $c=5$

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

$x=\frac{8\pm \sqrt{64-60}}{6}$

or, $x = \frac{8\pm \sqrt{4}}{6}$

or, $x = \frac{8\pm 2}{6}$

or, $x = \frac{10}{6}$ and $x = \frac{6}{6}$

or, $x=\frac{5}{3}$ and $x =1$

There are two distinct solutions given by $x = \frac{5}{3}$ and $x = 1$