Question

$$\begin{gathered} \frac{ax}{b}-\frac{by}{a}=a+b, \\ ax-by=2ab \end{gathered}$$

Answer 1

The given equations may be written as:
$\frac{ax}{b}-\frac{by}{a}-(a+b)=0$ ...(i)
$ax-by-2ab=0$ ...(ii)
Here, a₁ = $\frac{a}{b}$, b₁ = $\frac{-b}{a}$, c₁ = −(a + b), a₂ = a, b₂ = −b and c₂ = −2ab
By cross multiplication, we have:

∴ $\frac{x}{\left(-{\displaystyle \frac{b}{a}}\right)\times \left(-2ab\right)-\left(-b\right)\times \left(-\left(a+b\right)\right)}=\frac{y}{-\left(a+b\right)\times a-\left(-2ab\right)\times {\displaystyle \frac{a}{b}}}=\frac{1}{{\displaystyle \frac{a}{b}}\times \left(-b\right)-a\times \left(-{\displaystyle \frac{b}{a}}\right)}$
⇒ $\frac{x}{2b^2-b\left(a+b\right)}=\frac{y}{-a\left(a+b\right)+2a^2}=\frac{1}{-a+b}$
⇒ $\frac{x}{2b^2-ab-b^2}=\frac{y}{-a^2-ab+2a^2}=\frac{1}{-a+b}$
⇒ $\frac{x}{b^2-ab}=\frac{y}{a^2-ab}=\frac{1}{-\left(a-b\right)}$
⇒ $\frac{x}{-b\left(a-b\right)}=\frac{y}{a\left(a-b\right)}=\frac{1}{-\left(a-b\right)}$
⇒ $x=\frac{-b\left(a-b\right)}{-\left(a-b\right)}=b,y=\frac{a\left(a-b\right)}{-\left(a-b\right)}=-a$
Hence, x = b and y = −a is the required solution.