Question

$$\begin{gathered} \frac{x}{a}+\frac{y}{b}=(a+b), \\ \frac{x}{a^2}+\frac{y}{b^2}=2. \end{gathered}$$

Answer 1

The given equations may be written as:
$\frac{x}{a}+\frac{y}{b}-\left(a+b\right)=0$ ...(i)
$\frac{x}{a^2}+\frac{y}{b^2}-2=0$ ...(ii)
Here, a₁ = $\frac{1}{a}$, b₁ = $\frac{1}{b}$, c₁ = −(a + b), a₂ = $\frac{1}{a^2}$, b₂ = $\frac{1}{b^2}$ and c₂ = −2
By cross multiplication, we have:

∴ $\frac{x}{\left[\left({\displaystyle -2}\right)\times {\displaystyle \frac{1}{b}}-{\displaystyle \frac{1}{b^2}}\times \left(-\left(a+b\right)\right)\right]}=\frac{y}{\left[{\displaystyle \frac{1}{a^2}}\times \left(-\left(a+b\right)\right)-{\displaystyle \frac{1}{a}}\times \left(-2\right)\right]}=\frac{1}{\left[{\displaystyle \frac{1}{a{b}^2}}-{\displaystyle \frac{1}{a^{2}b}}\right]}$
⇒ $\frac{x}{\left({\displaystyle \frac{-2}{b}}+{\displaystyle \frac{a}{b^2}}+{\displaystyle \frac{b}{b^2}}\right)}=\frac{y}{\left({\displaystyle \frac{-1}{a}}-{\displaystyle \frac{b}{a^2}}+{\displaystyle \frac{2}{a}}\right)}=\frac{1}{{\displaystyle \frac{a-b}{a^2{b}^2}}}$
⇒ $\frac{x}{\left[{\displaystyle \frac{-2b+a+b}{b^2}}\right]}=\frac{y}{\left[{\displaystyle \frac{-a-b+2a}{a^2}}\right]}=\frac{1}{\left[{\displaystyle \frac{a-b}{a^2{b}^2}}\right]}$
⇒ $\frac{x}{{\displaystyle \frac{a-b}{b^2}}}=\frac{y}{{\displaystyle \frac{a-b}{a^2}}}=\frac{1}{{\displaystyle \frac{a-b}{a^2{b}^2}}}$

⇒ $x=\frac{{\displaystyle \frac{a-b}{b^2}}}{{\displaystyle \frac{a-b}{a^2{b}^2}}}=a^2,y=\frac{{\displaystyle \frac{a-b}{a^2}}}{{\displaystyle \frac{a-b}{a^2{b}^2}}}=b^2$
Hence, x = a² and y = b² is the required solution.