Question

$\frac{ax}{b} = \frac{1}{b} + \frac{bx-1}{a}(a \neq 0;b \neq 0 )$ brigado

Answer 1

$\text{Ol\'a!}\\\\\\\dfrac{ax}{b}=\dfrac{1}{b}+\dfrac{bx-1}{a}\\\\\dfrac{ax}{b}=\dfrac{1}{b}+\dfrac{bx}{a}-\dfrac{1}{a}\\\\\dfrac{ax}{b}-\dfrac{bx}{a}=\dfrac{1}{b}-\dfrac{1}{a}\\\\x\!\left(\dfrac{a}{b}-\dfrac{b}{a}\right)=\dfrac{a-b}{ba}\\\\x\!\left(\dfrac{a^2-b^2}{ba}\right)=\dfrac{a-b}{ba}\\\\x\,(a^2-b^2)=a-b\\\\x\left[(a-b)(a+b)\right]=a-b\\\\x\,(a+b)=1\\\\\boxed{\boxed{x=\dfrac{1}{a+b}}}$