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).

1 answer

To determine the number of solutions for the system of equations using the substitution method, we start by substituting one equation into the other.

We have:

  1. \( y = x^2 - 2x - 6 \) (Equation 1)
  2. \( y = -5x - 9 \) (Equation 2)

We can substitute Equation 2 into Equation 1. This means we will replace \( y \) in Equation 1 with the expression from Equation 2:

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

Next, we'll rearrange the equation by moving all terms to one side:

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

Combining like terms gives:

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

Now, we need to determine the number of solutions to the quadratic equation \( x^2 + 3x + 3 = 0 \). We can do this by calculating the discriminant (\( D \)):

\[ D = b^2 - 4ac \] where \( a = 1 \), \( b = 3 \), and \( c = 3 \).

\[ D = 3^2 - 4(1)(3) = 9 - 12 = -3 \]

Since the discriminant \( D \) is negative (\( D < 0 \)), this indicates that the quadratic equation has no real solutions.

Thus, the system of equations has 0 solutions.