# How do you solve 2/(x^2-16x+64)-8/(x-7)=(Ax+B)/(x^2-16x+63)?

##### 1 Answer
Dec 23, 2016

There is no solution.

#### Explanation:

$\frac{2}{{x}^{2} - 16 x + 64} - \frac{8}{x - 7}$

= $\frac{2 \left(x - 7\right) - 8 \left({x}^{2} - 16 x + 64\right)}{\left({x}^{2} - 2 \times 8 \times x + {8}^{2}\right) \left(x - 7\right)}$

= $\frac{2 x - 14 - 8 {x}^{2} + 128 x - 512}{{\left(x - 8\right)}^{2} \left(x - 7\right)}$

= $\frac{- 8 {x}^{2} + 130 x - 526}{{\left(x - 8\right)}^{2} \left(x - 7\right)}$

Further, $\frac{A x + B}{{x}^{2} - 16 x + 63}$

= $\frac{A x + B}{\left(x - 7\right) \left(x - 9\right)}$, and hence

$\frac{- 8 {x}^{2} + 130 x - 526}{{\left(x - 8\right)}^{2} \left(x - 7\right)} = \frac{A x + B}{\left(x - 7\right) \left(x - 9\right)}$

or $\frac{\left(x - 9\right) \left(- 8 {x}^{2} + 130 x - 526\right)}{{\left(x - 8\right)}^{2} \left(x - 7\right) \left(x - 9\right)} = \frac{{\left(x - 8\right)}^{2} \left(A x + B\right)}{{\left(x - 8\right)}^{2} \left(x - 7\right) \left(x - 9\right)}$

or $\left(x - 9\right) \left(- 8 {x}^{2} + 130 x - 526\right) = \left({x}^{2} - 16 x + 64\right) \left(A x + B\right)$

or $- 8 {x}^{3} + 130 {x}^{2} - 526 x + 72 {x}^{2} - 1170 x + 4734 = A {x}^{3} - 16 A {x}^{2} + 64 A x + B {x}^{2} - 16 B x + 64 B$

or $- 8 {x}^{3} + 202 {x}^{2} - 1696 x + 4734 = A {x}^{3} + \left(- 16 A + B\right) {x}^{2} + \left(64 A - 16 B\right) x + 64 B$

Hence, comparing like terms, we should have

$A = - 8$, $- 16 A + B = 202$

now putting $A = - 8$ gives $B = 74$

also then though $64 A - 16 B = - 512 - 1184 = - 1696$

and $64 B = 4736$

as the latter is not true

there is no solution.