Question

. If 1/x - 1/y = 7 and 1/(x ^ 2) - 1/(y ^ 2) = 21 what is the value of x?

Answer 1

To solve for \( x \) given the equations:

1. \( \frac{1}{x} - \frac{1}{y} = 7 \)
2. \( \frac{1}{x^2} - \frac{1}{y^2} = 21 \)

Let's first introduce substitutions for convenience. Let \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \). This transforms our equations into:

1. \( a - b = 7 \)
2. \( a^2 - b^2 = 21 \)

The second equation can be factored using the difference of squares:

\[
a^2 - b^2 = (a - b)(a + b) = 21
\]

Substituting \( a - b = 7 \) from the first equation into this factorization gives:

\[
7(a + b) = 21
\]

Dividing both sides by 7 results in:

\[
a + b = 3
\]

Now, we have two new equations:

1. \( a - b = 7 \)
2. \( a + b = 3 \)

We can solve this system of equations. Adding the two equations:

\[
(a - b) + (a + b) = 7 + 3
\]
\[
2a = 10 \quad \Rightarrow \quad a = 5
\]

Now, substituting \( a = 5 \) back into one of the equations to find \( b \):

\[
5 - b = 7 \quad \Rightarrow \quad -b = 7 - 5 \quad \Rightarrow \quad -b = 2 \quad \Rightarrow \quad b = -2
\]

Now we have found:

\[
a = 5 \quad \text{and} \quad b = -2
\]

Recall that \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \):

\[
\frac{1}{x} = 5 \quad \Rightarrow \quad x = \frac{1}{5}
\]
\[
\frac{1}{y} = -2 \quad \Rightarrow \quad y = -\frac{1}{2}
\]

Thus, the value of \( x \) is

\[
\boxed{\frac{1}{5}}.
\]