Question
Solve for x and y:
$a^2x + b^2y = c^2,$
$b^2x + a^2y = d^2$
$a^2x + b^2y = c^2,$
$b^2x + a^2y = d^2$
✓
Answer
$a^2x + b^2y = c^2...(i)$
$b^2x + a^2y = d^2...(ii)$
Multiplying (i) by $a^2$ and (ii) by $b^2$ and subtracting, we get
$\Rightarrow\text{a}^4\text{x} - \text{b}^4\text{x} = \text{a}^2\text{b}^2 - \text{b}^2\text{d}^2$
$\Rightarrow\text{x}=\frac{\text{a}^2\text{c}^2-\text{b}^2\text{d}^2}{\text{a}^4-\text{b}^4}$
Multiplying (i) by $b^2$ and (ii) by $a^2$ and subtracting, we get
$\Rightarrow\text{b}^4\text{y} - \text{a}^4\text{y} = \text{b}^2\text{c}^2 - \text{a}^2\text{d}^2$
$\Rightarrow\text{y}=\frac{\text{b}^2\text{c}^2-\text{a}^2\text{d}^2}{\text{b}^4-\text{a}^4}$
So, $\text{x}=\frac{\text{a}^2\text{c}^2-\text{b}^2\text{d}^2}{\text{a}^4-\text{b}^4}$ and $\text{y}=\frac{\text{b}^2\text{c}^2-\text{a}^2\text{d}^2}{\text{b}^4-\text{a}^4}$
$b^2x + a^2y = d^2...(ii)$
Multiplying (i) by $a^2$ and (ii) by $b^2$ and subtracting, we get
$\Rightarrow\text{a}^4\text{x} - \text{b}^4\text{x} = \text{a}^2\text{b}^2 - \text{b}^2\text{d}^2$
$\Rightarrow\text{x}=\frac{\text{a}^2\text{c}^2-\text{b}^2\text{d}^2}{\text{a}^4-\text{b}^4}$
Multiplying (i) by $b^2$ and (ii) by $a^2$ and subtracting, we get
$\Rightarrow\text{b}^4\text{y} - \text{a}^4\text{y} = \text{b}^2\text{c}^2 - \text{a}^2\text{d}^2$
$\Rightarrow\text{y}=\frac{\text{b}^2\text{c}^2-\text{a}^2\text{d}^2}{\text{b}^4-\text{a}^4}$
So, $\text{x}=\frac{\text{a}^2\text{c}^2-\text{b}^2\text{d}^2}{\text{a}^4-\text{b}^4}$ and $\text{y}=\frac{\text{b}^2\text{c}^2-\text{a}^2\text{d}^2}{\text{b}^4-\text{a}^4}$
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