# How do you solve and check for extraneous solutions in (x-4) / (x-1)= 10/ (x+7)?

Aug 30, 2015

$x = 9 \text{ }$ or $\text{ } x = - 2$

#### Explanation:

Every time you're dealing with rational expressions, you need to make sure that you exclude from the solution set any value of $x$ that can make the denominator(s) equal to zero.

$x - 1 \ne 0 \implies x \ne 1$

and

$x + 7 \ne 0 \implies x \ne - 7$

The common denominator for these two fractions will be $\left(x - 1\right) \left(x + 7\right)$, so multiply each side by the appropriate expression to get

$\frac{\left(x - 4\right) \cdot \left(x + 7\right)}{\left(x - 1\right) \left(x + 7\right)} = \frac{10 \cdot \left(x - 1\right)}{\left(x - 1\right) \left(x + 7\right)}$

This is equivalent to

$\left(x - 4\right) \left(x + 7\right) = 10 \left(x - 1\right)$

${x}^{2} + 3 x - 28 = 10 x - 10$

${x}^{2} - 7 x - 18 = 0$

Use the quadratic formula to find the two roots of the quadratic

${x}_{1 , 2} = \frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \cdot 1 \cdot \left(- 18\right)}}{2 \cdot 1}$

${x}_{1 , 2} = \frac{7 \pm \sqrt{121}}{2}$

${x}_{1 , 2} = \frac{7 \pm 11}{2} = \left\{\begin{matrix}{x}_{1} = \frac{7 + 11}{2} = 9 \\ {x}_{2} = \frac{7 - 11}{2} = - 2\end{matrix}\right.$

Both ${x}_{1}$ and ${x}_{2}$ satisfy the conditions $x \ne 1$ and $x \ne - 7$, which means that both will be valid solutions to the original equation.