To solve for \( x \) given the equations:
- \( \frac{1}{x} - \frac{1}{y} = 7 \)
- \( \frac{1}{x^2} - \frac{1}{y^2} = 21 \)
Let's first introduce substitutions for convenience. Let \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \). This transforms our equations into:
- \( a - b = 7 \)
- \( a^2 - b^2 = 21 \)
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:
- \( a - b = 7 \)
- \( a + b = 3 \)
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
\[ \boxed{\frac{1}{5}}. \]