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y+x^2=2x y+2x=4 find the solutions
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Solve the bottom one for 2x by subtracting y on both sides
Then you have both equations =2x
y=4-2x put it into y+x^2=2x (4-2x)+x^2=2x x^2-2x+4=2x x^2+4=0
So that means whatever they equal must equal
Or solve for y like cyclets did
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thanks
And also one thing there is an error -2x-2x=-4x
ops..sorry.. y=4-2x put it into y+x^2=2x (4-2x)+x^2=2x x^2-2x+4=2x x^2-4x+4=0 (x+2)(x-2)=0
(x-2)(x-2)=0
so would the points be 2,0 and -2,8
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the point is 0,0 to 2,0
wouldnt the x's be 2 and -2 though
I think he meant (x-2)(x-2)=0 which will get you one zero with multiplicity 2 x=2 So to find what y is just plug in 2 for x and solve for y (just use one of the equations given)
yes..thanks myininaya ~ appreciate that
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