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

Question 41

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

Solution

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), => $$\frac{x+y}{x-y}=\frac{7}{3}$$

=> Ans - (B)

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