Solve the system of equations: \{\begin{array}{l} x - y = 2 \\ y^2 - 2xy - x^2 = -28 \end{array} and \{\begin{array}{l} x + 2y^2 = 4 \\ x - y = 4 \end{array}

This problem presents two separate systems of equations. Let's solve each one.

System 1:

The first system is:

• \[ \begin{cases} x - y = 2 \\ y^2 - 2xy - x^2 = -28 \end{cases} \]

From the first equation, we can express x in terms of y: x = y + 2.

Substitute this into the second equation:

• \[ (y^2) - 2(y+2)y - (y+2)^2 = -28 \]
• \[ y^2 - (2y^2 + 4y) - (y^2 + 4y + 4) = -28 \]
• \[ y^2 - 2y^2 - 4y - y^2 - 4y - 4 = -28 \]
• \[ -2y^2 - 8y - 4 = -28 \]
• \[ -2y^2 - 8y + 24 = 0 \]
• Divide by -2:
• \[ y^2 + 4y - 12 = 0 \]

Factor the quadratic equation:

• \[ (y + 6)(y - 2) = 0 \]

This gives two possible values for y:

• y = -6 or y = 2

Now, find the corresponding values for x using x = y + 2:

• If y = -6, then x = -6 + 2 = -4.
• If y = 2, then x = 2 + 2 = 4.

So, the solutions for the first system are (-4, -6) and (4, 2).

System 2:

The second system is:

• \[ \begin{cases} x + 2y^2 = 4 \\ x - y = 4 \end{cases} \]

From the second equation, we can express x in terms of y: x = y + 4.

Substitute this into the first equation:

• \[ (y + 4) + 2y^2 = 4 \]
• \[ 2y^2 + y + 4 = 4 \]
• \[ 2y^2 + y = 0 \]

Factor out y:

• \[ y(2y + 1) = 0 \]

This gives two possible values for y:

• y = 0 or 2y + 1 = 0, which means y = -1/2

Now, find the corresponding values for x using x = y + 4:

• If y = 0, then x = 0 + 4 = 4.
• If y = -1/2, then x = -1/2 + 4 = -1/2 + 8/2 = 7/2.

So, the solutions for the second system are (4, 0) and (7/2, -1/2).

Final Answer: The solutions for the first system are (-4, -6) and (4, 2). The solutions for the second system are (4, 0) and (7/2, -1/2).