# Solve by factorization a/(ax-1)+b/(bx-1)=a+b

Solve by factorization
$\frac{a}{ax-1}+\frac{b}{bx-1}=a+b$
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Step 1: Given,
The expression $\frac{a}{ax-1}+\frac{b}{bx-1}=a+b$. We have to solve this by factorization.
Step 2: Calculation
$\frac{a}{ax-1}+\frac{b}{bx-1}=a+b$
$\frac{a\left(bx-1\right)+b\left(ax-1\right)}{\left(ax-1\right)\left(bx-1\right)}=a+b$
$\frac{abx-a+abx-b}{\left(ax-1\right)\left(bx-1\right)}=a+b$
$\frac{2abx-\left(a+b\right)}{\left(ax-1\right)\left(bx-1\right)}=a+b$
$2abx-\left(a+b\right)=\left(a+b\right)\left(ax-1\right)\left(bx-1\right)$
$2abx-\left(a+b\right)=\left(a+b\right)\left(ab{x}^{2}-ax-bx+1\right)$
$\left(a+b\right)ab{x}^{2}-{\left(a+b\right)}^{2}x+\left(a+b\right)-2abx+\left(a+b\right)=0$
$\left(a+b\right)ab{x}^{2}-\left[{\left(a+b\right)}^{2}-2ab\right]x+2\left(a+b\right)=0$
If we have a quadratic equation $a{x}^{2}+bx+c=0thenx=\frac{-b±\sqrt{{b}^{2}-4ac}}{2a}$
From the equation $\left(a+b\right)ab{x}^{2}-a+b2-2abx+2a+b=0$ we have
$x=\frac{\left[{\left(a+b\right)}^{2}-2ab\right]±\sqrt{{\left[{\left(a+b\right)}^{2}-2ab\right]}^{2}-8ab{\left(a+b\right)}^{2}}}{2ab\left(a+b\right)}$