Question

$$\begin{gathered} ax+by=(a-b), \\ bx-ay=(a+b). \end{gathered}$$

Answer 1

The given equations may be written as:
ax + by − (a − b) = 0 ...(i)
bx − ay − (a + b) = 0 ...(ii)
Here, a₁ = a, b₁ = b, c₁ = −(a − b), a₂ = b, b₂ = −a and c₂ = −(a + b)
By cross multiplication, we have:

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