Solving Simultaneous Equations Using Substitution Practice

To determine the number of solutions for the given linear-quadratic system of equations, we can use substitution. We have the following equations:

1. \( y = x^2 - 2x - 6 \) (quadratic equation)
2. \( y = -5x - 9 \) (linear equation)

Since both equations are equal to \( y \), we can set them equal to each other:

\[
x^2 - 2x - 6 = -5x - 9
\]

Next, we will rearrange the equation:

\[
x^2 - 2x + 5x - 6 + 9 = 0
\]

This simplifies to:

\[
x^2 + 3x + 3 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

In our equation, \( a = 1 \), \( b = 3 \), and \( c = 3 \). Plugging these values into the formula gives us:

\[
x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 3}}{2 \cdot 1}
\]
\[
x = \frac{-3 \pm \sqrt{9 - 12}}{2}
\]
\[
x = \frac{-3 \pm \sqrt{-3}}{2}
\]

Since we have a negative number under the square root \( (\sqrt{-3}) \), this indicates that there are no real solutions.

Thus, the system has **0 solutions**.