Solve system by eliminationy=x^2y=x+2SHOW YOUR WORK
Accepted Solution
A:
we have that y=x²----> equation 1 y=x+2-----> equation 2
multiply equation 1 by -1 -y=-x²
add equation 1 and equation 2 -y=-x² y=x+2 ------------ 0=-x²+x+2-------------> -x²+x+2=0-----> x²-x-2=0 Group terms that contain the same variable, and move the
constant to the opposite side of the equation(x²-x)=2 Complete
the square. Remember to balance the equation by adding the same constants
to each side (x²-x+0.5²)=2+0.5² Rewrite as perfect squares(x-0.5)²=2+0.5² (x-0.5)²=2.25-----> (x-0.5)=(+/-)√2.25-----> (x-0.5)=(+/-)1.5 x1=1.5+0.5-----> x1=2 x2=-1.5+0.5---- > x2=-1
for x=2 y=x²----> y=2²----> y=4 the point is (2,4)
for x=-1 y=x²----> y=(-1)²---> y=1 the point is (-1,1)
the answer is the solution of the system are the points (2,4) and (-1,1)