If $$x^2+y^2=29$$ and xy=10, where x>0,y>0,x>y then the value of $$\frac{x+y}{x-y}$$ is
Given : $$x^2+y^2=29$$ ---------(i)
Also, $$xy=10$$
=> $$2xy=20$$ ----------(ii)
Adding equations (i) and (ii),
=> $$x^2+y^2+2xy=29+20$$
=> $$(x+y)^2=49$$
=> $$x+y=\sqrt{49}=7$$ ----------(iii)
Similarly, subtracting equation (ii) from (i), we get :
=> $$(x-y)^2=29-20=9$$
=> $$x-y=3$$ -----------(iv)
Using equations (iii) and (iv), =>Â $$\frac{x+y}{x-y}=\frac{7}{3}$$
=> Ans - (B)
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