x-y=1 xy=12

To solve the system of equations given by:

1. \( x - y = 1 \)
2. \( xy = 12 \)

We can use substitution or elimination. Here, I'll use substitution:

From the first equation, we can express \( x \) in terms of \( y \):

\[ x = y + 1 \]

Now, substitute \( x \) into the second equation:

\[ (y + 1)y = 12 \]

Expanding this gives:

\[ y^2 + y = 12 \]

Rearranging to form a standard quadratic equation:

\[ y^2 + y - 12 = 0 \]

Next, we can factor this quadratic equation:

\[ (y - 3)(y + 4) = 0 \]

Setting each factor equal to zero gives us the solutions for \( y \):

1. \( y - 3 = 0 \) → \( y = 3 \)
2. \( y + 4 = 0 \) → \( y = -4 \)

Now we can find the corresponding values of \( x \) for each value of \( y \):

1. For \( y = 3 \): \[ x = y + 1 = 3 + 1 = 4 \] So one solution is \( (x, y) = (4, 3) \).

2. For \( y = -4 \): \[ x = y + 1 = -4 + 1 = -3 \] So the other solution is \( (x, y) = (-3, -4) \).

Thus, the two pairs \((x, y)\) that solve the system of equations are:

1. \((4, 3)\)
2. \((-3, -4)\)