Which point is the solution to the following system of equations? x^2+y^2=13 2x-y=4

Please make the separation of your two equations more obvious, as by writing only one equation per line:

x^2+y^2=13

2x-y=4

Solve the 2nd equation for y:  y = 2x-4.  Now substitute 2x-4 for y in the first equation:

x^2 + (2x-4)^2 = 13, or     4x^2 - 16x + 16 - 13 = 0.

Then 4x^2 - 16x + 3 = 0.  Solve by completing the square, as follows:

4(x^2 - 4x                       ) + 3  = 0
4(x^2 - 4x + 4 - 4            ) + 3  = 0
    4(x-2)^2 -16 +3                    = 0
     4(x-2)^2 = 13
Take the sqrt of both sides:  x-2 = plus or minus sqrt(13)

Then x = 2 plus or minus sqrt(13).  Each of these roots has an associated y-value.  How would you find it?  What are the 2 solutions of this system?

Hmmm i had this before , give me a sec

Y= -2x-1 the negative in front of the 2 is because the slope is negative and the -1 is because it’s down 1 on the y axis.