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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