We are given a system of two equations: $\frac{1}{2}y - \frac{1}{2}x = 1$ $x^2 + y^2 = 20$ We need to find the solutions $(x, y)$ that satisfy both equations.

1. Problem Description

We are given a system of two equations:

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

$x^2 + y^2 = 20$

We need to find the solutions $(x, y)$ that satisfy both equations.

2. Solution Steps

First, we can simplify the first equation:

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

Multiply both sides by 2:

$y - x = 2$

$y = x + 2$

Now, substitute this expression for $y$ into the second equation:

$x^2 + (x + 2)^2 = 20$

Expand the squared term:

$x^2 + (x^2 + 4x + 4) = 20$

Combine like terms:

$2x^2 + 4x + 4 = 20$

Subtract 20 from both sides:

$2x^2 + 4x - 16 = 0$

Divide both sides by 2:

$x^2 + 2x - 8 = 0$

Factor the quadratic equation:

$(x + 4)(x - 2) = 0$

So, $x = -4$ or $x = 2$.

If $x = -4$, then $y = x + 2 = -4 + 2 = -2$.

If $x = 2$, then $y = x + 2 = 2 + 2 = 4$.

Therefore, the solutions are $(-4, -2)$ and $(2, 4)$.

We can check our solutions:

For $(-4, -2)$:

$\frac{1}{2}(-2) - \frac{1}{2}(-4) = -1 + 2 = 1$

$(-4)^2 + (-2)^2 = 16 + 4 = 20$

For $(2, 4)$:

$\frac{1}{2}(4) - \frac{1}{2}(2) = 2 - 1 = 1$

$(2)^2 + (4)^2 = 4 + 16 = 20$

Both solutions satisfy the equations.

3. Final Answer

The solutions are $(-4, -2)$ and $(2, 4)$.