# How do you solve (3x-4)/(x^2-10x+21)-(x-8)/(x^2-18x+77)=(x-5)/(x^2-14x+33) and check for extraneous solutions?

Dec 24, 2016

$x = - 1 \mathmr{and} 15$..

#### Explanation:

graph{ y-(3x-4)/((x-7)(x-3))+(x-8)/((x-7)(x-11))+(x-5)/((x-11)(x-3))=0 [-15, 15, -20, 20]}
graph{y-(3x-4)/((x-7)(x-3))+(x-8)/((x-7)(x-11))+(x-5)/((x-11)(x-3))=0 [-50, 50, -1, 1]} Here,

$y = \frac{3 x - 4}{\left(x - 3\right) \left(x - 7\right)} - \frac{x - 8}{\left(x - 7\right) \left(x - 11\right)} - \frac{x - 5}{\left(x - 11\right) \left(x - 3\right)} = 0$.

This is not a problem, in limits. So, x is none of 3, 7 and 11.

( The graph for the limit problem is also inserted.)

Multiplying by $\left(x - 3\right) \left(x - 7\right) \left(x - 11\right)$,

$\left(3 x - 8\right) \left(x - 11\right) - \left(x - 8\right) \left(x - 3\right) - \left(x - 5\right) \left(x - 7\right) = 0$.

Upon simplification,

x^2-14x-15=0. Solving.

$x = - 1 \mathmr{and} 15$.

The two inserted graphs for the same function, on different scales,

are for depicting zeros for y that are solutions for the problem,

besides the indeterminate forms $\infty - \infty$, at x = 3, 7 and 11.