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?
