Please give the answer of the question Solve and find x and yax + by = 1 and bx - Maths - Pair of Linear Equations in Two Variables

Question

Please give the answer of the question

Solve and find x and y ax   +  
by   =   1   and   bx   +   ay  
=   2 ab a 2 + b 2

Ankush Jain · Accepted Answer

$a x + b y = 1 and b x + a y = \frac{2 a b}{a^{2} + b^{2}}$

The given equations can be rewritten as :

$a x + b y - 1 = 0 and b x + a y - \frac{2 a b}{a^{2} + b^{2}} = 0$

Using method of cross - multiplication, we get

$\frac{x}{\frac{- 2 a b^{2}}{a^{2} + b^{2}} + a} = \frac{- y}{\frac{- 2 a^{2} b}{a^{2} + b^{2}} + b} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{x}{\frac{- 2 a b^{2} + a^{3} + a b^{2}}{a^{2} + b^{2}}} = \frac{- y}{\frac{- 2 a^{2} b + a^{2} b + b^{3}}{a^{2} + b^{2}}} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{(a^{2} + b^{2}) x}{a^{3} - a b^{2}} = \frac{- y (a^{2} + b^{2})}{b^{3} - a^{2} b} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{(a^{2} + b^{2}) x}{a (a^{2} - b^{2})} = \frac{y (a^{2} + b^{2})}{b (a^{2} - b^{2})} = \frac{1}{a^{2} - b^{2}} \Rightarrow \frac{(a^{2} + b^{2}) x}{a (a^{2} - b^{2})} = \frac{1}{a^{2} - b^{2}} and \frac{y (a^{2} + b^{2})}{b (a^{2} - b^{2})} = \frac{1}{a^{2} - b^{2}} \Rightarrow x = \frac{a}{a^{2} + b^{2}} and y = \frac{b}{a^{2} + b^{2}}$

Please give the answer of the question Solve and find x and y ax + by = 1 and bx + ay = 2 ab a 2 + b 2

Please give the answer of the question

$Solve and find x and y ax + by = 1 and bx + ay = \frac{2 ab}{a^{2} + b^{2}}$