Question

solve the system of equations.
y=x2+13x+13
y=2x–17
Write the coordinates in exact form. Simplify all fractions and radicals.
,
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Answer 1

To solve the system of equations:

1. \( y = x^2 + 13x + 13 \) (Equation 1)
2. \( y = 2x - 17 \) (Equation 2)

We can set the two equations for \( y \) equal to each other since both are equal to \( y \):

\[
x^2 + 13x + 13 = 2x - 17
\]

Now, let's move all terms to one side of the equation:

\[
x^2 + 13x + 13 - 2x + 17 = 0
\]

This simplifies to:

\[
x^2 + 11x + 30 = 0
\]

Next, we will factor the quadratic equation. We need to find two numbers that multiply to \( 30 \) and add to \( 11 \). The numbers are \( 5 \) and \( 6 \).

Thus, we factor the equation as follows:

\[
(x + 5)(x + 6) = 0
\]

Setting each factor equal to zero gives:

\[
x + 5 = 0 \quad \Rightarrow \quad x = -5
\]
\[
x + 6 = 0 \quad \Rightarrow \quad x = -6
\]

Now we have two \( x \) values: \( x = -5 \) and \( x = -6 \). We will now substitute these back into one of the original equations to find the corresponding \( y \) values.

Let's use Equation 2 (\( y = 2x - 17 \)).

**For \( x = -5 \):**

\[
y = 2(-5) - 17 = -10 - 17 = -27
\]

The coordinates are \( (-5, -27) \).

**For \( x = -6 \):**

\[
y = 2(-6) - 17 = -12 - 17 = -29
\]

The coordinates are \( (-6, -29) \).

Putting everything together, the system of equations has the following solutions:

\[
\text{Coordinates: } (-5, -27) \text{ and } (-6, -29)
\]