# How do you solve (x-4)/(x-8)+(x-5)/(x-7)=(x+143)/(x^2-15x+56) and check for extraneous solutions?

Oct 30, 2016

The equation has no solution.

#### Explanation:

First, you must factor the denominators as possible to determine what the Least Common Denominator of the equation is. In this equation, both denominators on the left of the equals sign cannot be factored.

$\frac{x - 4}{x - 8} + \frac{x - 5}{x - 7} = \frac{x + 143}{{x}^{2} - 15 x + 56}$

$\frac{x - 4}{x - 8} + \frac{x - 5}{x - 7} = \frac{x + 143}{\left(x - 7\right) \left(x - 8\right)}$

So the LCD of this equation is $\left(x - 7\right) \left(x - 8\right)$. Multiply each term by this LCD, remembering that you can cross-reduce like factors from the denominator before multiplying.

$\left(x - 7\right) \left(x - 8\right) \left(\frac{x - 4}{x - 8}\right) + \left(x - 7\right) \left(x - 8\right) \left(\frac{x - 5}{x - 7}\right) = \left(x - 7\right) \left(x - 8\right) \left(\frac{x + 143}{\left(x - 7\right) \left(x - 8\right)}\right)$

In the first term, the $\left(x - 8\right)$ in the LCD and in the denominator of the original fraction cross-reduce each other. In the second term, the $\left(x - 7\right)$ in the LCD and in the denominator of the original fraction cross-reduce each other. In the last term, the full LCD and the full denominator of the original fraction cross-reduce each other. The result is:

$\left(x - 7\right) \left(x - 4\right) + \left(x - 8\right) \left(x - 5\right) = x + 143$
${x}^{2} - 4 x - 7 x + 28 + {x}^{2} - 5 x - 8 x + 40 = x + 143$
$2 {x}^{2} - 24 x + 68 = x + 143$
$2 {x}^{2} - 24 x - x + 68 = x - x + 143$
$2 {x}^{2} - 25 x + 68 = 143$
$2 {x}^{2} - 25 x + 68 - 143 = 143 - 143$
$2 {x}^{2} - 25 x - 75 = 0$

Now, factor the resulting quadratic equation.

$\left(2 x - 15\right) \left(x - 5\right) = 0$

Now, use the Zero Product Property to find the two possible values of $x$.

$2 x - 15 = 0$
$2 x - 15 + 15 = 0 + 15$
$2 x = 15$
$\frac{2 x}{2} = \frac{15}{2}$
$x = 7.5$

And

$x - 5 = 0$
$x - 5 + 5 = 0 + 5$
$x = 5$

Now, to check for extraneous solutions, each of these possible solutions must be tested in the original equation to see if they make it true. We will begin by first replacing every $x$ in the original equation with the first possible solution, $7.5$.

$\frac{7.5 - 4}{7.5 - 8} + \frac{7.5 - 5}{7.5 - 7} = \frac{7.5 + 143}{{7.5}^{2} - 15 \left(7.5\right) + 56}$

$\frac{3.5}{-} 0.5 + \frac{2.5}{0.5} = \frac{150.5}{56.25 - 112.5 + 56}$

$- 7 + 5 = \frac{150.5}{-} 0.25$
$- 2 \ne - 602$

Now test the second possible solution, $5$, in the same way.

$\frac{5 - 4}{5 - 8} + \frac{5 - 5}{5 - 7} = \frac{5 + 143}{{5}^{2} - 15 \left(5\right) + 56}$

$\frac{1}{-} 3 + \frac{0}{-} 2 = \frac{148}{25 - 75 + 56}$

$\frac{- 1}{3} - 0 = \frac{148}{6}$

$\frac{- 1}{3} \ne 24 \frac{2}{3}$

Since neither solution makes the original equation true, they are both extraneous, and the equation has no solution.