# How do you solve (x-4)/(x-7)+(x-5)/(x-6)=(x+164)/(x^2-13x+42) and check for extraneous solutions?

May 22, 2018

${x}_{1} = \frac{15}{2}$ or ${x}_{2} = - \frac{7}{2}$

#### Explanation:

We have
$\left(x - 7\right) \left(x - 6\right) = {x}^{2} - 13 x + 42$
so it must be
$x \ne 7 , x \setminus \ne 6$
Multiplying by$\left(x - 6\right) \left(x - 7\right)$
we get
$\left(x - 4\right) \left(x - 6\right) + \left(x - 5\right) \left(x - 7\right) = x + 164$
expanding and collecting
$2 {x}^{2} - 22 x + 59 = x + 164$
$2 {x}^{2} - 22 x - 105 = 0$
${x}^{2} - \frac{23}{2} x - \frac{105}{2} = 0$
using the quardatic formula we get
${x}_{1} = 15$
${x}_{2} = - \frac{7}{2}$

Aug 14, 2018

$x = - \frac{7}{2} \text{ }$ or $\text{ } x = 15$

#### Explanation:

Given:

$\frac{x - 4}{x - 7} + \frac{x - 5}{x - 6} = \frac{x + 164}{{x}^{2} - 13 x + 42}$

Note that:

${x}^{2} - 13 x + 42 = \left(x - 7\right) \left(x - 6\right)$

So multiplying both sides of the given equation by $\left(x - 7\right) \left(x - 6\right)$ we get:

$\left(x - 4\right) \left(x - 6\right) + \left(x - 5\right) \left(x - 7\right) = x + 164$

Multiplied out that becomes:

${x}^{2} - 10 x + 24 + {x}^{2} - 12 x + 35 = x + 164$

That is:

$2 {x}^{2} - 22 x + 59 = x + 164$

Subtracting $x + 164$ from both sides, that becomes:

$2 {x}^{2} - 23 x - 105 = 0$

Using an AC method look for a pair of factors of $A C = 2 \cdot 105 = 210$ which differ by $B = 23$.

The pair $30 , 7$ works.

Use this pair to split the middle term and factor by grouping:

$0 = 2 {x}^{2} - 23 x - 105$

$\textcolor{w h i t e}{0} = \left(2 {x}^{2} - 30 x\right) + \left(7 x - 105\right)$

$\textcolor{w h i t e}{0} = 2 x \left(x - 15\right) + 7 \left(x - 15\right)$

$\textcolor{w h i t e}{0} = \left(2 x + 7\right) \left(x - 15\right)$

Hence:

$x = - \frac{7}{2} \text{ }$ or $\text{ } x = 15$

Note that neither of these values cause a zero denominator in the given rational equation, so they are both valid solutions.