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
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=2$
$\frac{\text{bx}+\text{ay}}{\text{ab}}=2$
$ b x+a y=2 a b \ldots(1) $
$ a x-b y=\left(a^2-b^2\right) ...(2)$
Multiplying (1) by b and (2) by a
$ \Rightarrow b^2 x+b a y=2 a b^2 \ldots(3) $
$ \Rightarrow a^2 x-b a y=a\left(a^2-b^2\right) ...(4)$
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+a^2\right)=a b^2+a^3$
$ x\left(b^2+a^2\right)=a\left(b^2+a^2\right)$
$\text{x}=\frac{\text{a}\big(\text{b}^2+\text{a}^2\big)}{\big(\text{b}^2+\text{a}^2\big)}=\text{a}$
Putting $x = a$ in $(1)$, we get
$b × a + ay = 2ab$
$ay = 2ab - ab$
$ay = ab$ or $y = b$
$\therefore$ Solution is $x = a, y = b$
$\frac{\text{bx}+\text{ay}}{\text{ab}}=2$
$ b x+a y=2 a b \ldots(1) $
$ a x-b y=\left(a^2-b^2\right) ...(2)$
Multiplying (1) by b and (2) by a
$ \Rightarrow b^2 x+b a y=2 a b^2 \ldots(3) $
$ \Rightarrow a^2 x-b a y=a\left(a^2-b^2\right) ...(4)$
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+a^2\right)=a b^2+a^3$
$ x\left(b^2+a^2\right)=a\left(b^2+a^2\right)$
$\text{x}=\frac{\text{a}\big(\text{b}^2+\text{a}^2\big)}{\big(\text{b}^2+\text{a}^2\big)}=\text{a}$
Putting $x = a$ in $(1)$, we get
$b × a + ay = 2ab$
$ay = 2ab - ab$
$ay = ab$ or $y = b$
$\therefore$ Solution is $x = a, y = b$
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