Answer
The solutions are $(-3,-6)$ and $(3,6)$.
Work Step by Step
The given system is
$y=\frac{1}{3}x^2+2x-3$ ...... (1)
$y=2x$ ...... (2)
Graph each equation.
The point of intersections are $A=(-3,-6)$ and $B=(3,6)$.
Check: $(x,y)=(-3,-6)$
Equation (1).
$\Rightarrow y=\frac{1}{3}x^2+2x-3$
$\Rightarrow -6=\frac{1}{3}(-3)^2+2(-3)-3$
$\Rightarrow -6=3-6-3$
$\Rightarrow -6=3-9$
$\Rightarrow -6=-6$
True.
Equation (2).
$\Rightarrow y=2x$
$\Rightarrow -6=2(-3)$
$\Rightarrow -6=-6$
True.
Check: $(x,y)=(3,6)$
$\Rightarrow y=\frac{1}{3}x^2+2x-3$
$\Rightarrow 6=\frac{1}{3}(3)^2+2(3)-3$
$\Rightarrow 6=3+6-3$
$\Rightarrow 6=9-3$
$\Rightarrow 6=6$
True.
Equation (2).
$\Rightarrow y=2x$
$\Rightarrow 6=2(3)$
$\Rightarrow 6=6$
True.
Hence, the solutions are $(-3,-6)$ and $(3,6)$.
