ax+by=1 and bx+ay= 2ab/(a^2+b^2) Solve by Cross Multiplication Method

Cross Multiplication Method

ax+by=1

⟹ ax+by-1=0 —–(i)

bx+ay = $\frac{2ab}{a^2 + b^2}$

⟹ b(a²+b²)x+a(a²+b²)y -2ab = 0

⟹ (ba²+b³)x+(a³+ab²)y-2ab=0 —- (ii)

ax+by-1=0 —–(i)

(ba²+b³)x+(a³+ab²)y-2ab=0 —- (ii)

Applying Cross Multiplication Method in (i) and (ii) We Get,

$\frac{x}{b( – 2ab) + a^3 + ab^2}$ = $\frac{y}{ – (ba^2 + b^3) + 2a^2b}$=$\frac{1}{a(a^3 + ab^2) – b(ba^2 + b^3)}$

⟹$\frac{x}{ – 2ab^2 + a^3 + ab^2}$ = $\frac{y}{2a^2b – ba^2 – b^3}$ = $\frac{1}{a^4 + a^2b^2 – b^2a^2 – b^4} $

$\Rightarrow \frac{x}{a^3 – ab^2}$ = $\frac{y}{ba^2 – b^3}$ = $\frac{1}{a^{4 -}b^4}$

$\Rightarrow \frac{x}{a\left(a^2 – b^2\right)}$ = $\frac{y}{b(a^2 – b^2)}$ = $\frac{1}{(a^2 + b^2)(a^2 – b^2)}$

$\Rightarrow \frac{x}{a}$ = $\frac{y}{b}$ = $\frac{1}{(a^2 + b^2)}$

$\therefore x = \frac{a}{(a^2 + b^2)}$ ,y = $\frac{b}{(a^2 + b^2)}$

∴ Required Solutions are x = $\frac{a}{(a^2 + b^2)}$ ,y = $\frac{b}{(a^2 + b^2)}$

ax+by=1 —-(i)

bx+ay= 2ab/(a^2+b^2) —-(ii)

Multiplying Equation (i) with a and Equation (ii) with b we get,

a²x+aby = a —- (iii)

b²x +aby = $\frac{2ab^2}{a^2 + b^2}$ —-(iv)

Now Subtracting (iv) from (iii) 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 = $\frac{b}{(a^2 + b^2)}$