Solve for x and y: bx a - ay b +a+b=0, bx-ay+2ab=0 - Vidyadip

Question

Solve for $x$ and $y$:
$\frac{\text{bx}}{\text{a}}-\frac{\text{ay}}{\text{b}}+\text{a}+\text{b}=0,$
$\text{bx}-\text{ay}+\text{2ab}=0$

✓

Answer

$\frac{\text{bx}}{\text{a}}-\frac{\text{ax}}{\text{b}}+\text{a}+\text{b}=0$
By taking $L.C.M$., we get
$\frac{\text{b}^2\text{x}-\text{a}^2\text{y}+\text{a}^2\text{b}+\text{b}^2\text{a}}{\text{ab}}=0$
$b^2x - a^2y = -a^2b - b^2a ...(1)$
$bx - ay = - 2ab ...(2)$
Multiplying $(1)$ by $1$ and $(2)$ by a
$b^2x - a^2y = -a^2b - b^2a ...(3)$
$abx - a^2b = - 2a^2b ...(4)$
Subtracting $(3)$ from $(4)$
$(ab - b^2)x = -2a^2b + a^2b + ab^2$
$b(a - b)x = -a^2b + ab^2 = -ab(a - b)$
$\therefore\ \text{x}=\frac{-\text{ab}(\text{a}-\text{b})}{\text{b}(\text{a}-\text{b})}$
$x = -a$
Putting $x = -a$, in $(1)$, we get
$b^2(-a) - a^2y = -a^2b - b^2a$
$-ab^2 - a^2y = -a^2b - b^2a$
$-a^2y = -a^2b - b^2a + ab^2$
$-a^2y = -a^2b$
$\Rightarrow\text{y}=\frac{-\text{a}^2\text{b}}{-\text{a}^2}=\text{b}$
$\therefore$ solution is $x = -a, y = b$

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