To solve the system of equations given by
- \( y = -22x - 8 \)
- \( y = x^2 - 27x - 4 \)
we can set the right-hand sides of the equations equal to each other since both expressions equal \( y \):
\[ -22x - 8 = x^2 - 27x - 4 \]
Next, we rearrange the equation to set it to zero:
\[ 0 = x^2 - 27x - 4 + 22x + 8 \]
Combining like terms gives:
\[ 0 = x^2 - 5x + 4 \]
Next, we will solve the quadratic equation \( x^2 - 5x + 4 = 0 \) using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -5 \), and \( c = 4 \):
\[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} \] \[ x = \frac{5 \pm \sqrt{25 - 16}}{2} \] \[ x = \frac{5 \pm \sqrt{9}}{2} \] \[ x = \frac{5 \pm 3}{2} \]
This gives us two solutions for \( x \):
- \( x = \frac{5 + 3}{2} = \frac{8}{2} = 4 \)
- \( x = \frac{5 - 3}{2} = \frac{2}{2} = 1 \)
Now we will substitute these \( x \) values back into one of the original equations to find the corresponding \( y \) values. We will use the first equation \( y = -22x - 8 \).
For \( x = 4 \): \[ y = -22(4) - 8 = -88 - 8 = -96 \] So one solution is \( (4, -96) \).
For \( x = 1 \): \[ y = -22(1) - 8 = -22 - 8 = -30 \] So another solution is \( (1, -30) \).
Thus, the coordinates of the solutions in exact form are:
\[ \boxed{(4, -96) \text{ and } (1, -30)} \]