Question : If $x^2+y^2=29$ and $xy=10$, where $x>0,y>0$ and $x>y$. Then the value of $\frac{x+y}{x-y}$ is:
Option 1: $- \frac{7}{3}$
Option 2: $\frac{7}{3}$
Option 3: $\frac{3}{7}$
Option 4: $-\frac{3}{7}$
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Correct Answer: $\frac{7}{3}$
Solution :
Given: $x^2+y^2=29$, $xy=10$
$x^2+y^2=29$
⇒ $x^2+y^2+20=29+20$ (adding 20 on both sides)
⇒ $x^2+y^2+2×10=49$
⇒ $x^2+y^2+2×xy=49$ (as $xy=10$)
⇒ $(x+y)^2=7^2$
⇒ $x+y=7$
Now,
⇒ $x^2+y^2-20=29-20$ (subtracting 20 on both sides)
⇒ $x^2+y^2-2×xy=9$ (as $xy=10$)
⇒ $(x-y)^2=3^2$
⇒ $x-y=3$
In the same way, we get $x-y=3$
So, $\frac{x+y}{x-y}=\frac{7}{3}$
Hence, the correct answer is $\frac{7}{3}$.
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