Chapter A.6, Problem 78AYU

To determine

To solve: $\frac{5}{x+4}=4+\frac{3}{x-2}$

Answer to Problem 78AYU

$x=1, -\frac{5}{2}$

Explanation of Solution

$\frac{5}{x+4}=4+\frac{3}{x-2}$

$\frac{5\left(x+4\right)\left(x-2\right)}{x+4}=4\left(x+4\right)\left(x-2\right)+\frac{3\left(x+4\right)\left(x-2\right)}{x-2}$   (Multiply $\left(x+4\right)\left(x-2\right)$ on both sides)

$5\left(x-2\right)=\left(4x+16\right)\left(x-2\right)+3\left(x+4\right)$

$5x-10=4x^2+16x-8x-32+3x+12$

$4x^2+6x-10=0$

$a{x}^2 + bx + c = 0$ Compare to standard form.

With $a=4, b=6$, and $c=-10$, the discriminant is $b^2-4ac={\left(6\right)}^2-4\left(4\right)\left(-10\right)=196$.

Since $b^2-4ac>0$, there are two real solutions. Use the quadratic formula with $a=4, b=6, c=-10$ and $b^2-4ac=196$.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(6\right)\pm \sqrt{196}}{2\left(4\right)}=\frac{-6\pm 14}{8}$

$x=\frac{-6+14}{8}=1$ and $x=\frac{-6-14}{8}=-\frac{20}{8}=-\frac{5}{2}$

Therefore $x=1, -\frac{5}{2}$.