X-y=4 Xy=12

You have a system of equations:

1. \( x - y = 4 \)
2. \( xy = 12 \)

Let's solve these equations step by step.

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

\[ x = y + 4 \]

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

\[ (y + 4)y = 12 \]

Expanding this gives:

\[ y^2 + 4y = 12 \]

Now, move all terms to one side of the equation:

\[ y^2 + 4y - 12 = 0 \]

Next, we can solve this quadratic equation using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

In this case, \( a = 1 \), \( b = 4 \), and \( c = -12 \). Plugging in these values:

\[ y = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-12)}}{2 \cdot 1} \] \[ y = \frac{-4 \pm \sqrt{16 + 48}}{2} \] \[ y = \frac{-4 \pm \sqrt{64}}{2} \] \[ y = \frac{-4 \pm 8}{2} \]

This gives us two possible solutions for \( y \):

1. \( y = \frac{4}{2} = 2 \)
2. \( y = \frac{-12}{2} = -6 \)

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

1. If \( y = 2 \): \[ x = 2 + 4 = 6 \]

2. If \( y = -6 \): \[ x = -6 + 4 = -2 \]

Thus, the solutions to the system of equations are:

1. \( (x, y) = (6, 2) \)
2. \( (x, y) = (-2, -6) \)

So the final answer is:

• The pairs \( (x, y) \) are \( (6, 2) \) and \( (-2, -6) \).