# Solve (x+3)/(x+2) + (x+4)/(x+3) = (x+5)/(x+4) - (x+6)/(x+5)?

Nov 1, 2017

The solution of:

$\frac{x + 3}{x + 2} \textcolor{red}{-} \frac{x + 4}{x + 3} = \frac{x + 5}{x + 4} - \frac{x + 6}{x + 5}$

is $x = - \frac{7}{2}$

#### Explanation:

Suppose the question should be:

$\frac{x + 3}{x + 2} \textcolor{red}{-} \frac{x + 4}{x + 3} = \frac{x + 5}{x + 4} - \frac{x + 6}{x + 5}$

Making common denominators on the left hand side and on the right hand side, this becomes:

$\frac{\left(x + 3\right) \left(x + 3\right) - \left(x + 2\right) \left(x + 4\right)}{\left(x + 2\right) \left(x + 3\right)} = \frac{\left(x + 5\right) \left(x + 5\right) - \left(x + 4\right) \left(x + 6\right)}{\left(x + 4\right) \left(x + 5\right)}$

Multiplying out the numerators, we get:

$\frac{\left({x}^{2} + 6 x + 9\right) - \left({x}^{2} + 6 x + 8\right)}{\left(x + 2\right) \left(x + 3\right)} = \frac{\left({x}^{2} + 10 x + 25\right) - \left({x}^{2} + 10 x + 24\right)}{\left(x + 4\right) \left(x + 5\right)}$

Most of the terms in the numerator cancel, to give us:

$\frac{1}{\left(x + 2\right) \left(x + 3\right)} = \frac{1}{\left(x + 4\right) \left(x + 5\right)}$

Taking the reciprocal of both sides, this becomes:

$\left(x + 2\right) \left(x + 3\right) = \left(x + 4\right) \left(x + 5\right)$

which multiplies out as:

${x}^{2} + 5 x + 6 = {x}^{2} + 9 x + 20$

Subtracting ${x}^{2} + 5 x + 20$ from both sides, this becomes:

$- 14 = 4 x$

Dividing both sides by $2$ and transposing, we get:

$x = - \frac{7}{2}$

Nov 1, 2017

In the form given this resolves to a typical quartic with approximate roots:

${x}_{1} \approx - 9.4400$

${x}_{2} \approx - 0.28158$

${x}_{3} \approx - 2.6392 + 4.5893 i$

${x}_{4} \approx - 2.6392 - 4.5893 i$

#### Explanation:

Assuming the question is correct as given...

Given:

$\frac{x + 3}{x + 2} + \frac{x + 4}{x + 3} = \frac{x + 5}{x + 4} - \frac{x + 6}{x + 5}$

Subtract the right hand side from the left to get:

$\frac{x + 3}{x + 2} + \frac{x + 4}{x + 3} - \frac{x + 5}{x + 4} + \frac{x + 6}{x + 5} = 0$

Transposing and multiplying both sides by $\left(x + 2\right) \left(x + 3\right) \left(x + 4\right) \left(x + 5\right)$ this becomes:

$0 = {\left(x + 3\right)}^{2} \left(x + 4\right) \left(x + 5\right) + \left(x + 2\right) {\left(x + 4\right)}^{2} \left(x + 5\right) - \left(x + 2\right) \left(x + 3\right) {\left(x + 5\right)}^{2} + \left(x + 2\right) \left(x + 3\right) \left(x + 4\right) \left(x + 6\right)$

$\textcolor{w h i t e}{0} = \left({x}^{4} + 15 {x}^{3} + 83 {x}^{2} + 201 x + 180\right) + \left({x}^{4} + 15 {x}^{3} + 82 {x}^{2} + 192 x + 160\right) - \left({x}^{4} + 15 {x}^{3} + 81 {x}^{2} + 185 + 150\right) + \left({x}^{4} + 15 {x}^{3} + 80 {x}^{2} + 180 x + 144\right)$

$\textcolor{w h i t e}{0} = 2 {x}^{4} + 30 {x}^{3} + 164 {x}^{2} + 573 x + 149$

This is a typical quartic, with two real irrational zeros and two non-real complex zeros.

It is possible but very messy to solve algebraically. Using a numerical method such as Durand-Kerner we find approximate solutions:

${x}_{1} \approx - 9.4400$

${x}_{2} \approx - 0.28158$

${x}_{3} \approx - 2.6392 + 4.5893 i$

${x}_{4} \approx - 2.6392 - 4.5893 i$

See https://socratic.org/s/aKtpkf7J for more details.