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Kris5
solve the following system of equations graphically. 2x^2-4x=y+1 x+y=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 \[y=y~~hence~~2x^2-4x-1 = -x+1\] \[2x^2-4x-1 = -x +1\]
\[2x^2-4x-1 = -x +1\]\[2x^2-4x-1-1 = -x\]\[2x^2-4x-2+x =0\]\[2x^2-3x-1 = 0\] Then solve for x by using the quadratic formula, you'll find two x's, since it's the system is something like this: |dw:1335839518743:dw|