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Mathematics 7 Online
OpenStudy (anonymous):

how do you solve y=x^2-11x+24 y=x-3 Solve each system

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 :)

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

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

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?

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

OpenStudy (radar):

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