Respuesta :
If a solution(s) exists y=y so we can say:
x^2-3x=-2x+2 add 2x to both sides
x^2-x=2 subtract 2 from both sides
x^2-x-2=0 factor
(x-2)(x+1)=0
So x=-1 and 2, using y=-2x+2 we find:
y(-1)=4 and y(2)=-2
So the two solutions occur at the points:
(-1,4) and (2,-2)
x^2-3x=-2x+2 add 2x to both sides
x^2-x=2 subtract 2 from both sides
x^2-x-2=0 factor
(x-2)(x+1)=0
So x=-1 and 2, using y=-2x+2 we find:
y(-1)=4 and y(2)=-2
So the two solutions occur at the points:
(-1,4) and (2,-2)
Answer:it has 2 solutions
Step-by-step explanation: Just took the quiz
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