Answer
The system has no real solutions.
Work Step by Step
The given system is
$y=-3x^2+x+2$ ...... (1)
$y=x+4$ ...... (2)
Subtract equation (2) from equation (1).
$\Rightarrow y-y=-3x^2+x+2-(x+4)$
Clear the parentheses.
$\Rightarrow y-y=-3x^2+x+2-x-4$
Simplify.
$\Rightarrow 0=-3x^2-2$
Add $3x^2$ to each side.
$\Rightarrow 0+3x^2=-3x^2-2+3x^2$
Simplify.
$\Rightarrow 3x^2=-2$
Divide each side by $3$.
$\Rightarrow \frac{3x^2}{3}=-\frac{2}{3}$
Simplify.
$\Rightarrow x^2=-\frac{2}{3}$
The square of a real number cannot be negative.
The equation has no real solutions.
Hence, the system has no real solutions.
