Question

Solve for x and y:
$$\begin{gathered} a^{2}x+b^{2}y=c^2 \\ {b}^{2}x+a^{2}y=d^2 \end{gathered}$$

Answer 1

The given equations are
$$\begin{gathered} a^{2}x+b^{2}y=c^2.....\left(i\right) \\ {b}^{2}x+a^{2}y=d^2.....\left(ii\right) \end{gathered}$$
Multiplying (i) by a² and (ii) by b² and subtracting, we get
$$\begin{gathered} a^{4}x-b^{4}x=a^2{c}^2-b^2{d}^2 \\ \Rightarrow x=\frac{a^2{c}^2-b^2{d}^2}{a^4-b^4} \end{gathered}$$
Now, multiplying (i) by b² and (ii) by a² and subtracting, we get
$$\begin{gathered} b^{4}y-a^{4}y=b^2{c}^2-a^2{d}^2 \\ \Rightarrow y=\frac{b^2{c}^2-a^2{d}^2}{b^4-a^4} \end{gathered}$$

Hence, $x=\frac{a^2{c}^2-b^2{d}^2}{a^4-b^4}\mathrm{and}y=\frac{b^2{c}^2-a^2{d}^2}{b^4-a^4}$.