To determine the number of solutions to the system of equations given by \( y = x^2 + 6x + 7 \) and \( y = -3x + 19 \), we can set the two equations equal to each other:
\[ x^2 + 6x + 7 = -3x + 19 \]
Next, we will rearrange the equation to set it to zero:
\[ x^2 + 6x + 3x + 7 - 19 = 0 \]
Combining like terms gives:
\[ x^2 + 9x - 12 = 0 \]
Now we will solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our case, \( a = 1 \), \( b = 9 \), and \( c = -12 \).
First, we need to calculate the discriminant \( b^2 - 4ac \):
\[ b^2 - 4ac = 9^2 - 4 \cdot 1 \cdot (-12) \] \[ = 81 + 48 \] \[ = 129 \]
Since the discriminant is positive (\( 129 > 0 \)), this indicates that there are two distinct real solutions for \( x \).
Thus, the system of equations has two solutions, so the answer is:
2
1 answer