Solve the system of equations: $x^2 + y^2 = 10$ $y - x = 2$

1. Problem Description

Solve the system of equations:

$x^2 + y^2 = 10$

$y - x = 2$

2. Solution Steps

From the second equation, we can express $y$ in terms of $x$:

$y = x + 2$

Substitute this expression for $y$ into the first equation:

$x^2 + (x + 2)^2 = 10$

Expand the equation:

$x^2 + (x^2 + 4x + 4) = 10$

Combine like terms:

$2x^2 + 4x + 4 = 10$

Subtract 10 from both sides:

$2x^2 + 4x - 6 = 0$

Divide both sides by 2:

$x^2 + 2x - 3 = 0$

Factor the quadratic equation:

$(x + 3)(x - 1) = 0$

Solve for $x$:

$x = -3$ or $x = 1$

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

If $x = -3$, then $y = -3 + 2 = -1$

If $x = 1$, then $y = 1 + 2 = 3$

So, the solutions are $(-3, -1)$ and $(1, 3)$.

3. Final Answer

The solutions are $(-3, -1)$ and $(1, 3)$.