Solving Simultaneous Equations Using Substitution Practice

Question

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Question
Use the substitution method to determine whether the linear-quadratic system y=x2−2x−6
, y=−5x−9
has 0, 1, or 2 solutions.(1 point)
This system has
solution(s).
Remaining Attempts : 3
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Answer 1

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.