Solve the system 1/x + 1/y = -1 and 3/x - 2/y = 7.

Step-by-step solution

1. 1

   Substitute new variables for the reciprocals. Clearing denominators would produce quadratic cross terms $xy$. Instead let

   $a = \frac{1}{x}, \qquad b = \frac{1}{y}$

   The system becomes perfectly linear:

   $a + b = -1, \qquad 3a - 2b = 7$

   Both $x$ and $y$ must be non-zero for the original equations to make sense, so nothing is lost.

2. 2

   Eliminate one unknown. From the first equation $a = -1 - b$. Substituting into the second:

   $3(-1 - b) - 2b = 7 \quad\Longrightarrow\quad -3 - 5b = 7 \quad\Longrightarrow\quad b = -2$

3. 3

   Back-substitute for a.

   $a = -1 - (-2) = 1$

4. 4

   Undo the substitution. Since $a = \tfrac1x = 1$ we get $x = 1$, and since $b = \tfrac1y = -2$ we get

   $y = -\frac{1}{2}$

   Both are non-zero, so both are admissible.

5. 5

   Check in the original equations. $\tfrac{1}{1} + \tfrac{1}{-1/2} = 1 - 2 = -1$ and $\tfrac{3}{1} - \tfrac{2}{-1/2} = 3 + 4 = 7$. Both hold exactly.