Question

If \(\frac{x+y}{x-y}+\frac{x-y}{x+y}=\frac{10}{3}\),then \(\frac xy=\)

• A. \(\pm2\)
• B. \(\pm1\)
• C. \(\pm4\)
• D. Not Possible to find out

Answer 1

The Correct Option is A

To solve the equation: \(\frac{x+y}{x-y}+\frac{x-y}{x+y}=\frac{10}{3}\), let's follow these steps.

First, let \(a=\frac{x+y}{x-y}\) and \(b=\frac{x-y}{x+y}\). We know from the problem statement that \(a+b=\frac{10}{3}\).

Notice that \(a=\frac{1}{b}\) because if you multiply \(\frac{x+y}{x-y}\) by \(\frac{x-y}{x+y}\), you get 1:

\(\left(\frac{x+y}{x-y}\right)\left(\frac{x-y}{x+y}\right)=\frac{(x+y)(x-y)}{(x-y)(x+y)}=1\).

This implies \(b=\frac{1}{a}\).

Thus, we have:

\(a+\frac{1}{a}=\frac{10}{3}\)

Multiply throughout by \(a\) to clear the fraction:

\(a^2+1=\frac{10}{3}a\)

Rearrange this into a standard quadratic equation:

\(3a^2-10a+3=0\)

Use the quadratic formula \(a=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\) where \(a=3\), \(b=-10\), and \(c=3\):

\(a=\frac{-(-10)\pm\sqrt{(-10)^2-4\cdot3\cdot3}}{2\cdot3}\)

\(a=\frac{10\pm\sqrt{100-36}}{6}\)

\(a=\frac{10\pm\sqrt{64}}{6}\)

\(a=\frac{10\pm8}{6}\)

This gives \(a=\frac{18}{6}=3\) or \(a=\frac{2}{6}=\frac{1}{3}\).

Now, \(a=\frac{x+y}{x-y}=3\) leads to \(\frac{b}{a}=1\) implying \(\frac{x-y}{x+y}=\frac{1}{3}\).

Equating both conditions:

\(\frac{x+y}{x-y}=3\) and \(\frac{x-y}{x+y}=\frac{1}{3}\)

Multiply the two equations to verify identity:

\(\left(\frac{x+y}{x-y}\right)\left(\frac{x-y}{x+y}\right)=3\cdot\frac{1}{3}=1\), thus verified.

The ratio \(x:y\) is required:

\(x+y=3(x-y)\)

\(x+y=3x-3y\)

\(4y=2x\)

\(\frac{x}{y}=2\) or \(-2\) leading to \(\pm2\).

Therefore, the correct answer is \(\pm2\).