Question
Solve for x and y:
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=2,$
$\text{ax}-\text{by}=\text{a}^2-\text{b}^2$
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=2,$
$\text{ax}-\text{by}=\text{a}^2-\text{b}^2$
✓
Answer
$\frac{x}{a}+\frac{y}{b}=2$
$\frac{b x+a y}{a b}=2 b x+a y=2 a b \ldots$
$a x-b y=\left(a^2-b^2\right) \ldots(2)$
Multiplying (1) by b and (2) by a
$\Rightarrow b^2 x+\text { bay }=2 a b^2 \ldots(3)$
$\Rightarrow a^2 x-b a y=a\left(a^2-b^2\right) \ldots(4) \text { Adding (3)and (4), we get } b^2 x+a^2 x=2 a b^2+a\left(a^2-b^2\right) x\left(b^2+a^2\right)=2 a b^2+a^3-a b^2 x\left(b^2\right.$
$\left.+a^2\right)=a b^2+a^3 x\left(b^2+a^2\right)=a\left(b^2+a^2\right)$
$x=\frac{a\left(b^2+a^2\right)}{\left(b^2+a^2\right)}=a \text { Putting } x=a \text { in (1), we get } b \times a+a y=2 a b a y=2 a b-a b a$ $y=a b \text { or } y=b$
$\therefore$ Solution is $x=a, y=b$
$\frac{b x+a y}{a b}=2 b x+a y=2 a b \ldots$
$a x-b y=\left(a^2-b^2\right) \ldots(2)$
Multiplying (1) by b and (2) by a
$\Rightarrow b^2 x+\text { bay }=2 a b^2 \ldots(3)$
$\Rightarrow a^2 x-b a y=a\left(a^2-b^2\right) \ldots(4) \text { Adding (3)and (4), we get } b^2 x+a^2 x=2 a b^2+a\left(a^2-b^2\right) x\left(b^2+a^2\right)=2 a b^2+a^3-a b^2 x\left(b^2\right.$
$\left.+a^2\right)=a b^2+a^3 x\left(b^2+a^2\right)=a\left(b^2+a^2\right)$
$x=\frac{a\left(b^2+a^2\right)}{\left(b^2+a^2\right)}=a \text { Putting } x=a \text { in (1), we get } b \times a+a y=2 a b a y=2 a b-a b a$ $y=a b \text { or } y=b$
$\therefore$ Solution is $x=a, y=b$
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