# How do you solve (x-6)/(x-4)+(x-7)/(x-3)=(x+111)/(x^2-7x+12) and check for extraneous solutions?

Oct 1, 2016

$x = - 2 \frac{1}{2} \text{ or } x = 13$

#### Explanation:

In an equation with fractions, you can get rid of the fractions by multiplying every term by the LCD.

$\frac{x - 6}{x - 4} + \frac{x - 7}{x - 3} = \frac{x + 111}{{x}^{2} - 7 x + 12} \text{ } \leftarrow$ factor the denominator

$\frac{x - 6}{x - 4} + \frac{x - 7}{x - 3} = \frac{x + 111}{\left(x - 4\right) \left(x - 3\right)} \text{ } L C D = \textcolor{red}{\left(x - 4\right) \left(x - 3\right)}$

In the interests of space, let color(red)((x-4) = Aand (x-3) = B

$\frac{\textcolor{red}{\cancel{A} \cdot B} \times \left(x - 6\right)}{\cancel{x - 4}} + \frac{\textcolor{red}{A \cdot \cancel{B} \times} \left(x - 7\right)}{\cancel{x - 3}} = \frac{\cancel{\textcolor{red}{A \cdot B \times}} \left(x + 111\right)}{\cancel{\left(x - 4\right) \left(x - 3\right)}}$

$\left(x - 3\right) \left(x - 6\right) + \left(x - 4\right) \left(x - 7\right) = x + 111 \text{ } \leftarrow$ no fractions!

${x}^{2} - 9 x + 18 + {x}^{2} - 11 x + 28 = x + 111 \text{ } \leftarrow$ no brackets!

$2 {x}^{2} - 20 x - x + 46 - 111 = 0 \text{ } \leftarrow$ make the quadratic = 0

$2 {x}^{2} - 21 x - 65 = 0 \text{ } \leftarrow$ simplify and then find factors

$\left(2 x + 5\right) \left(x - 13\right) = 0 \text{ } \leftarrow$ solve with each factor =0

If $2 x + 5 = 0 \text{ } \rightarrow x = - 2 \frac{1}{2}$
If $x - 13 = 0 \text{ } \rightarrow x = 13$