ax+by=1 bx+ay=(a+b)^2/a^2+b^2-1 Solve the Pair of Simultaneous Linear Equations by Elimination Method
ax+by=1 –(i)
$bx + ay = \frac{(a + b)^2}{a^2 + b^2} – 1$ –(ii)
From equation (ii) we get,
bx + ay = $\frac{(a + b)^2 – a^2 – b^2}{a^2 + b^2}$
⟹ bx + ay =$\frac{a^2 + 2ab + b^2 – a^2 – b^2}{a^2 + b^2}$
⟹ bx + ay = $\frac{2ab}{a^2 + b^2}$ –(iii)
Multiplying Equation (i) with a and Equation (iii) with b we get,
a2x+aby = a –(iv)
b2x +aby = $\frac{2ab^2}{a^2 + b^2}$ –(v)
Now Subtracting (v) from (iv) we get,
(a2x+aby ) – (b2x +aby)=a -$\frac{2ab^2}{a^2 + b^2}$
⟹ a2x +aby -b2x -aby =${\mathrm{a}} – \frac{2{\mathrm{ab}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
⟹ x(a2-b2) =$\frac{{\mathrm{a}}\left({\mathrm{a}}^2 + {\mathrm{b}}^2\right) – 2{\mathrm{ab}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
⟹ x(a2-b2) =$\frac{{\mathrm{a}}^3 + {\mathrm{ab}}^2 – 2{\mathrm{ab}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
⟹ x(a2-b2) = $\frac{{\mathrm{a}}^3 – {\mathrm{ab}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
⟹ x(a2-b2) = $\frac{{\mathrm{a}}\left({\mathrm{a}}^2 – {\mathrm{b}}^2\right)}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
⟹ x=$\frac{{\mathrm{a}}}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
Substituting the Value of x in Equation (i) We Get,
${\mathrm{a}}\left(\frac{{\mathrm{a}}}{{\mathrm{a}}^2 + {\mathrm{b}}^2}\right) + {\mathrm{by}}$ = 1
$\Rightarrow \frac{{\mathrm{a}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2} + {\mathrm{by}}$ = 1
$\Rightarrow {\mathrm{by}}$ = 1 – $\frac{{\mathrm{a}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
$\Rightarrow {\mathrm{by}}$ = $\frac{{\mathrm{a}}^2 + {\mathrm{b}}^2 – a^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
$\Rightarrow {\mathrm{by}}$ = $\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
$\Rightarrow {\mathrm{y}}$ = $\frac{{\mathrm{b}}}{{\mathrm{a}}^2 + {\mathrm{b}}^2}$
∴ Required Solutions are x = $\frac{a}{(a^2 + b^2)}$ ,y = $\frac{b}{(a^2 + b^2)}$