Solve equation (1) for y
y-2x = 3
y-2x+2x = 3+2x
y = 2x+3
Plug that into equation (2)
y - 3 = x^2
2x+3 - 3 = x^2
2x = x^2
Get everything to one side
2x = x^2
2x-2x = x^2-2x
0 = x^2-2x
x^2-2x = 0
Now factor and use the zero product property to find the solutions for x
x^2-2x = 0
x(x-2) = 0
x = 0 or x-2 = 0
x = 0 or x = 2
If x = 0, then y is...
y = 2x+3
y = 2(0)+3
y = 3
So (x,y) = (0,3) is one solution of the system. This is one point where the graphs intersect.
If x = 2, then y is...
y = 2x+3
y = 2(2)+3
y = 4+3
y = 7
So (x,y) = (2,7) is another solution to the system. This is another point where the graphs intersect.
See the attached image for a visual. The two curves cross at point A and point B
A = (0,3)
B = (2,7)
y-2x = 3
y-2x+2x = 3+2x
y = 2x+3
Plug that into equation (2)
y - 3 = x^2
2x+3 - 3 = x^2
2x = x^2
Get everything to one side
2x = x^2
2x-2x = x^2-2x
0 = x^2-2x
x^2-2x = 0
Now factor and use the zero product property to find the solutions for x
x^2-2x = 0
x(x-2) = 0
x = 0 or x-2 = 0
x = 0 or x = 2
If x = 0, then y is...
y = 2x+3
y = 2(0)+3
y = 3
So (x,y) = (0,3) is one solution of the system. This is one point where the graphs intersect.
If x = 2, then y is...
y = 2x+3
y = 2(2)+3
y = 4+3
y = 7
So (x,y) = (2,7) is another solution to the system. This is another point where the graphs intersect.
See the attached image for a visual. The two curves cross at point A and point B
A = (0,3)
B = (2,7)