Question

If $x=a+\frac{1}{b+}\frac{1}{b+}\frac{1}{a+}\frac{1}{a+}....,$

$y=b+\frac{1}{a+}\frac{1}{a+}\frac{1}{b+}\frac{1}{b+}....,$.

Then show that $(a{b}^2+a+b)x-(a^{2}b+a+b)y=a^2-b^2$.

Answer 1

$x-a=\frac{Given,1}{b+}\frac{1}{a+}\frac{1}{b+}\frac{1}{a+}....$

$=\frac{1}{b+}\frac{1}{b+y-b}$

$\Rightarrow x-a=\frac{y}{by+1}$

$\Rightarrow byx+x-aby-a=y⟶1$

Similarly, $y-b=\frac{1}{a+}\frac{1}{x}=\frac{x}{ax+1}$

$\Rightarrow axy+y-(ab+1)x-b=0⟶2$

$\Rightarrow ax\left(1\right)-bx\left(2\right):-$

$\Rightarrow ax-(a^{2}b+a)y-by+(a{b}^2+b)x=a^2-b^2$

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