solve for x in the following Quadratic equations

Question

I am not able to solve this problem for my son. Is ther any error in the question itself. $\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$. Thanks in advance..

Answer 1

Your equation can be rewritten as $$\frac{1}{x+5}-\frac{1}{x+4}=\frac{1}{x+2}-\frac{1}{x+1}$$ Or $$\frac{1}{(x+4)(x+5)}=\frac{1}{(x+1)(x+2)}$$. Can you take it from here?

Answer 2

$$ \frac{1}{x+1}+\frac{1}{x+5}=\frac{1}{x+2}+\frac{1}{x+4} $$ Notice that $x\ne\{-5,-4,-2,-1\}$. $$ \frac{2x+6}{(x+1)(x+5)}=\frac{2x+6}{(x+2)(x+4)} $$ One solution will be $2x+6=0\implies x=-3$.
Second solution will be $$ (x+1)(x+5)=(x+2)(x+4)\\ x^2+6x+5=x^2+6x+8\\ $$ which gives you $5=8$, so only solution is $x=-3$.

Answer 3

$\dfrac{1}{(x+1)} + \dfrac{1}{(x+5)}=\dfrac{1}{(x+2)}+\dfrac{1}{(x+4)}$

$\implies \dfrac{1}{(x+5)}-\dfrac{1}{(x+4)}=\dfrac{1}{(x+2)}-\dfrac{1}{(x+1)} $

$\implies \dfrac{x+4-x-5}{(x+4)(x+5)}=\dfrac{x+1-x-2}{(x+1)(x+2)}$

$\implies (x+4)(x+5)=(x+1)(x+2)$

$\implies x^2+9x+20=x^2+3x+2$

$\implies 9x-3x+20-2=0$

Or,$$x=-3$$