Originally Posted by
Rapha It's really hard to read this.
Do you mean
$\displaystyle \frac{1}{3x} + \frac{1}{x+1} = \frac{1}{x^2+x}$ ?
If so multiply by x^2+x
$\displaystyle \frac{x^2+x}{3x} + \frac{x^2+x}{x+1} = 1$
$\displaystyle \frac{x^2+x}{3x} + \frac{x(x+1)}{x+1} = 1$
$\displaystyle \frac{x^2+x}{3x} + x = 1$
$\displaystyle \frac{x^2}{3x}+\frac{x}{3x} + x = 1$
$\displaystyle \frac{x}{3}+\frac{1}{3} + x = 1$
did you mean $\displaystyle \frac{1}{3}x+\frac{1}{x}+1 = \frac{1}{x^2+x}$ ?
multiply by x^2+x
$\displaystyle \frac{1}{3}x*(x^2+x)+\frac{x^2+x}{x}+x^2+x = 1$
$\displaystyle \frac{1}{3}x*(x^2+x)+\frac{x(x+1)}{x}+x^2+x = 1$
$\displaystyle \frac{1}{3}x*(x^2+x)+x+1+x^2+x = 1$
...
Rapha