Solve the following simultaneous equations: 2x + y = 5 and x2 + y2 = 25.

$$2x + y = 5 \quad (*1)$$ $$x^2 + y^2 = 25 \quad (*2)$$

From equation (*1):

$$y = 5 - 2x$$

$$x^2 + (5^2 - 2(5)(2x) + (2x)^2) = 25$$ $$x^2 + (25 - 20x + 4x^2) = 25$$

Combine like terms:

$$5x^2 - 20x + 25 = 25$$

Subtract 25 from both sides:

$$5x^2 - 20x = 0$$

Factor out $5x$:

$$5x(x - 4) = 0$$

This gives two possible solutions for $x$:

$$5x = 0 \implies x = 0$$ $$x - 4 = 0 \implies x = 4$$

For $x = 0$:

$$y = 5 - 2(0)$$ $$y = 5 - 0$$ $$y = 5$$

One solution is $(x, y) = (0, 5)$.

For $x = 4$:

$$y = 5 - 2(4)$$ $$y = 5 - 8$$ $$y = -3$$

The second solution is $(x, y) = (4, -3)$.

The solutions to the simultaneous equations are $x=0, y=5 \text{ and } x=4, y=-3$.