Question

Solve the system of equations.
y=
–
22x–8
y=x2–27x–4
Write the coordinates in exact form. Simplify all fractions and radicals.
,
,

Answer 1

To solve the system of equations given by

1. \( y = -22x - 8 \)
2. \( y = x^2 - 27x - 4 \)

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 \( x \):

1. \( x = \frac{5 + 3}{2} = \frac{8}{2} = 4 \)
2. \( x = \frac{5 - 3}{2} = \frac{2}{2} = 1 \)

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:

\[
\boxed{(4, -96) \text{ and } (1, -30)}
\]