Question

ax+by=a² ,bx+ay= b²​

Answer 1

Given

$\longrightarrow ax+by=a^2\quad\quad\dots(1)$

$\longrightarrow bx+ay=b^2\quad\quad\dots(2)$

Adding (1) and (2),

$\longrightarrow (ax+by)+(bx+ay)=a^2+b^2$

$\longrightarrow ax+by+bx+ay=a^2+b^2$

$\longrightarrow a(x+y)+b(x+y)=a^2+b^2$

$\longrightarrow (x+y)(a+b)=a^2+b^2$

$\longrightarrow x+y=\dfrac{a^2+b^2}{a+b}\quad\quad\dots(1)$

Subtracting (2) from (1),

$\longrightarrow (ax+by)-(bx+ay)=a^2-b^2$

$\longrightarrow ax+by-bx-ay=a^2-b^2$

$\longrightarrow a(x-y)-b(x-y)=(a-b)(a+b)$

$\longrightarrow (x-y)(a-b)=(a-b)(a+b)$

$\longrightarrow x-y=a+b\quad\quad\dots(4)$

Adding (3) and (4),

$\longrightarrow (x+y)+(x-y)=\dfrac{a^2+b^2}{a+b}+a+b$

$\longrightarrow x+y+x-y=\dfrac{a^2+b^2+(a+b)^2}{a+b}$

$\longrightarrow 2x=\dfrac{2a^2+2b^2+2ab}{a+b}$

$\longrightarrow\underline{\underline{x=\dfrac{a^2+b^2+ab}{a+b}}}$

Subtracting (4) from (3),

$\longrightarrow (x+y)-(x-y)=\dfrac{a^2+b^2}{a+b}-(a+b)$

$\longrightarrow x+y-x+y=\dfrac{a^2+b^2-(a+b)^2}{a+b}$

$\longrightarrow 2y=-\dfrac{2ab}{a+b}$

$\longrightarrow\underline{\underline{y=-\dfrac{ab}{a+b}}}$

Thus solved!