You have to equal the two #Y#s, meaning their values as well or you can find the value of the first #x# and then plug it in the second equation. There are many ways to solve this.
#y=x+3# and #y=2x^2#
#y=y=>x+3=2x^2=>2x^2-x-3=0#
You can use any tools you know to solve this quadratic equation but as for me, I will use #Delta#
#Delta=b^2-4ac#, with #a=2#, #b=-1# and #c=-3#
#Delta=(-1)^2-4(2)(-3)=25=> sqrt Delta=+-5#
#x_1=(-b+sqrt Delta)/(2a)# and #x_2=(-b-sqrt Delta)/(2a)#
#x_1=(1+5)/(4)=6/4=3/2# and #x_2=(1-5)/(4)=-1#
#x_1=3/2# and #x_2=-1#
To find #y#, all you have to do is to plug the #x# values in either of the two equations. I will plug in both just to show you that it does not matter which one you chose.
With the first equation #y=x+3#
For #x=3/2=>y=3/2+3=(3+6)/2=9/2#
For #x=-1=>y=-1+3=2#
With the second equation #y=2x^2#
For #x=3/2=>y=2(3/2)^2=1 color (red) cancel 2(9/(2 color (red) cancel4))=9/2#
For #x=-1=>y=2(-1)^2=2#
Therefore, your solution is #(3/2,9/2)# and #(-1,2)#
Hope this helps :)
