Question

solve the following system of equations graphically. 2x^2-4x=y+1
x+y=1

Answer 1

Ok, the first one is a parabola, if you put y to one side, you'll get
\[2x^2-4x-1=y\]
the second one is a linear function, if you put y to one side, you'll get \[y=-x+1\]
By using the comparaison method
\[y=y~~hence~~2x^2-4x-1 = -x+1\]
\[2x^2-4x-1 = -x +1\]

Answer 2

\[2x^2-4x-1 = -x +1\]\[2x^2-4x-1-1 = -x\]\[2x^2-4x-2+x =0\]\[2x^2-3x-1 = 0\]

Then solve for x by using the quadratic formula, you'll find two x's, since it's the system is something like this:

Answer 3

oh alright ty :)