Answer
The solutions are $(2,0)$ and $(6,4)$.
Work Step by Step
The given system is
$y=\frac{1}{2}x^2-3x+4$ ...... (1)
$y=x-2$ ...... (2)
Graph each equation.
The point of intersections are $A=(2,0)$ and $B=(6,4)$.
Check: $(x,y)=(2,0)$
Equation (1).
$\Rightarrow y=\frac{1}{2}x^2-3x+4$
$\Rightarrow 0=\frac{1}{2}(2)^2-3(2)+4$
$\Rightarrow 0=2-6+4$
$\Rightarrow 0=6-6$
$\Rightarrow 0=0$
True.
Equation (2).
$\Rightarrow y=x-2$
$\Rightarrow 0=2-2$
$\Rightarrow 0=0$
True.
Check: $(x,y)=(6,4)$
Equation (1).
$\Rightarrow y=\frac{1}{2}x^2-3x+4$
$\Rightarrow 4=\frac{1}{2}(6)^2-3(6)+4$
$\Rightarrow 4=18-18+4$
$\Rightarrow 4=4$
True.
Equation (2).
$\Rightarrow y=x-2$
$\Rightarrow 4=6-2$
$\Rightarrow 4=4$
True.
Hence, the solutions are $(2,0)$ and $(6,4)$.
