Solve the system of equations: xy = 4 and x+y = 5.

We are given two equations:

1. $xy = 4$
2. $x+y = 5$

From equation (2), we can express $y$ in terms of $x$:

$$y = 5 - x$$

Substitute $y = 5 - x$ into equation (1):

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

Distribute $x$:

$$5x - x^2 = 4$$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$$x^2 - 5x + 4 = 0$$

Factor the quadratic equation. We need two numbers that multiply to 4 and add to -5. These numbers are -1 and -4.

$$(x - 1)(x - 4) = 0$$

This gives two possible values for $x$:

$$x - 1 = 0 \implies x = 1$$ $$x - 4 = 0 \implies x = 4$$

Case 1: If $x = 1$

$$y = 5 - 1 = 4$$

Case 2: If $x = 4$

$$y = 5 - 4 = 1$$

The solutions for the system of equations are:

$$(x,y) = (1,4) \text{ or } (4,1)$$