Mathematics
7 Online
OpenStudy (anonymous):
how do you solve y=x^2-11x+24 y=x-3 Solve each system
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OpenStudy (anonymous):
Alright. Let's plug in x-3 where you see y.
What would your equation look like?
OpenStudy (anonymous):
x-3=1x^2-11x+24?
OpenStudy (anonymous):
Great :D
Now, let's get everything on one side and have zero on the other side. What do you have then?
OpenStudy (anonymous):
0=1x^2-10x+27?
OpenStudy (anonymous):
-12x instead of -10x, since you're subtracting x from both sides :)
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OpenStudy (anonymous):
oooh
OpenStudy (anonymous):
So you have x^2-10x+27. Now, can you factor this?
OpenStudy (anonymous):
you mean -12?
OpenStudy (anonymous):
-12, my bad XD
OpenStudy (anonymous):
lol
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OpenStudy (anonymous):
LOL xD
OpenStudy (anonymous):
Let's try that again:
See if you can factor x^2-12x+27 :)
OpenStudy (anonymous):
-9&-3
OpenStudy (anonymous):
(x-3)(x-9) :3
So x would be 3 and 9..
OpenStudy (anonymous):
the answer should be 3,0 and 9,6 though
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OpenStudy (radar):
It will be, when you finish the problem and solve for y.
OpenStudy (anonymous):
Oh, yup XD
OpenStudy (anonymous):
hooow?
OpenStudy (radar):
You have solved for x.
The problem states y = x - 3
Substitute your x values and solve for the two y values.
OpenStudy (anonymous):
oh its that easy?
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OpenStudy (anonymous):
uh i got -12 so thats not right
OpenStudy (radar):
Yes. y = 9 -3 = 6 (9,6)
y = 3 - 3 = 0 (3,0)
OpenStudy (anonymous):
i got -9 & -3
OpenStudy (radar):
(x-9)(x-3)= 0 solve getting x=9 and 3
OpenStudy (anonymous):
ok ok...how do you do y=2x^2+x+4 y=-x^2-x+9
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OpenStudy (radar):
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