How many solutions does the system have?
y=−2x+2
y=x^2−3x

Y=-2x+2 and y=x^2-3x.

If there are solutions the x and y values are equal so we can say y=y:

x^2-3x=-2x+2  add 2x to both sides

x^2-x=2  subtract 2 from both sides

x^2-x-2=0  now factor...

x^2+x-2x-2=0

x(x+1)-2(x+1)=0

(x-2)(x+1)=0

So there are two solutions, when x=-1 and 2

Using y=-2x+2 we can find the corresponding y values for the solutions...

y(-1)=-2(-1)+2=2+2=4, y(2)=-2(2)+2=-4+2=-2 so the two solutions are the points:

(-1,4) and (2,-2)