I need help finding the solutions of the system. i fan and Medal 1. y=x²+3x-4 y=2x+2 2. y+x²-2x-2 y=4x+5
some typos here?
\[y=x^2+3x-4\] \[y=2x+2\] set \[2x+2=x^2+3x-4\] and solve for\(x\)
start by setting equal to zero, then solve the quadratic equation \[2x+2=x^2+3x-4\] \[x^2+x-6=0\] factor \[(x-2)(x+3)=0\] set each factor equal to zero and solve
second one is similar, set \[x^2-2x-2=4x+5\]
ok x-2=0 +2 +2 x=2 x+3=0 -3 -3 x=3 So no slutions?
@amistre64 ^ is this what he was talking about?
(or she)
that is what he is talking about yes, but why would you conclude NO solutions?
because they aren't the same number?
the solution to a curve and a line will have either 2 solutions|dw:1365970152736:dw| 1 solution |dw:1365970174462:dw| or no real solutions |dw:1365970197889:dw|
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