Notice how easily we can isolate x in the equation y^2=x+3. We can subtract 3 from both sides of the equation to reveal that color(blue)(x=y^2-3). With this, we can replace color(blue)x in the second equation, color(blue)x-2y=12, with the value that we know color(blue)x is equal to: color(blue)(y^2-3).
This gives us: " "color(blue)x-2y=12" "=>" "(color(blue)(y^2-3))-2y=12
Continuing to solve the resulting equation, we obtain y^2-2y-15=0, which can be factored into (y-5)(y+3)=0 and solved giving two solutions for y, which are that color(red)(y=5) and color(green)(y=-3).
Each of the values of y can be plugged into either one of the two equations to find their corresponding values of x. I will choose the first equation, although you will receive the same results should you choose to input your values of y into the second.
Solving for x when color(red)(y=5), we obtain:
color(red)y^2=x+3" "=>" "color(red)5^2=x+3" "=>" "color(red)(x=22)
Since color(red)(x=22) and color(red)(y=5), we have a solution point at (color(red)22,color(red)5).
And for when color(green)(y=-3), we obtain:
color(green)y^2=x+3" "=>" "(color(green)(-3))^2=x+3" "=>" "color(green)(x=6)
Thus there is also a solution point at (color(green)6,color(green)(-3)).
Graphing the two equations should reveal a sideways parabola with a line intersecting it at the points (22,5) and (6,-3):
graph{(y^2-x-3)(x-2y-12)=0 [-5.14, 30.9, -7.66, 10.36]}
