Question
Solve for $x$ and $y$
$\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=0,$
$\text{ax}+\text{by}=(\text{a}^2+\text{b}^2)$
$\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=0,$
$\text{ax}+\text{by}=(\text{a}^2+\text{b}^2)$
✓
Answer
$\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=0$
$\Rightarrow\text{x}=\frac{\text{ay}}{\text{b}}\ \dots(\text{i})$
$\text{ax}+\text{by}=(\text{a}^2+\text{b}^2)\ \dots(\text{ii})$
Substituting $(i)$ in $(ii)$, we get
$\text{a}\Big(\frac{\text{ay}}{\text{b}}\Big)+\text{by}=(\text{a}^2+\text{b}^2)$
$\Rightarrow\Big(\frac{\text{a}^2\text{y}}{\text{b}}\Big)+\text{by}=(\text{a}^2+\text{b}^2)$
$\Rightarrow\text{a}^2\text{y}+\text{b}^2\text{y}=(\text{a}^2\text{b}+\text{b}^3)$
$\Rightarrow\text{y}(\text{a}^2+\text{b}^2)=\text{b}(\text{a}^2+\text{b}^2)$
$\Rightarrow\text{y}=\text{b}$
Substituting in $(i)$, we get $x = a$
So, $x = a$ and $y = b.$
$\Rightarrow\text{x}=\frac{\text{ay}}{\text{b}}\ \dots(\text{i})$
$\text{ax}+\text{by}=(\text{a}^2+\text{b}^2)\ \dots(\text{ii})$
Substituting $(i)$ in $(ii)$, we get
$\text{a}\Big(\frac{\text{ay}}{\text{b}}\Big)+\text{by}=(\text{a}^2+\text{b}^2)$
$\Rightarrow\Big(\frac{\text{a}^2\text{y}}{\text{b}}\Big)+\text{by}=(\text{a}^2+\text{b}^2)$
$\Rightarrow\text{a}^2\text{y}+\text{b}^2\text{y}=(\text{a}^2\text{b}+\text{b}^3)$
$\Rightarrow\text{y}(\text{a}^2+\text{b}^2)=\text{b}(\text{a}^2+\text{b}^2)$
$\Rightarrow\text{y}=\text{b}$
Substituting in $(i)$, we get $x = a$
So, $x = a$ and $y = b.$
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