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# Simultaneous Equations help!

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1. I'm doing simultaneous equations and I've got to (2y-12)(y+2), how do I work out y at this point?
2. (Original post by lemo****69)
I'm doing simultaneous equations and I've got to (2y-12)(y+2), how do I work out y at this point?
You can't, the quadratic not equal to anything.
3. (Original post by lemo****69)
I'm doing simultaneous equations and I've got to (2y-12)(y+2), how do I work out y at this point?
Hi please post the original question, thanks!
4. Is the quadratic you have there equal to 0?
5. Please Post the original question.
6. (Original post by lemo****69)
I'm doing simultaneous equations and I've got to (2y-12)(y+2), how do I work out y at this point?
Make sure you write down that it's equal to zero.
(2y-12)(y+2) = 0

2y = 12
y = 6

or

y = -2

So the answer is y = 6, y = -2
7. x+y = 4
x*2+y*2 = 40
solve this

I substituted y into it so I had:
2y*2-8y+16 = 40
2y*2-8y-24 = 0

I factorised and got:
(2y-12)(y+4)
8. (Original post by Namita Gurung)
Make sure you write down that it's equal to zero.
(2y-12)(y+2) = 0

2y = 12
y = 6

or

y = -2

So the answer is y = 6, y = -2
Thanks!
9. (Original post by lemo****69)
x+y = 4
x*2+y*2 = 40
solve this

I substituted y into it so I had:
2y*2-8y+16 = 40
2y*2-8y-24 = 0

I factorised and got:
(2y-12)(y+4)
(2y-12)(y+4) = 0

So you know for that to be true, one (or both) of those brackets must equal zero.

2y -12 = 0
Y + 4 = 0

Can you take it from here?
10. (Original post by lemo****69)
x+y = 4
x*2+y*2 = 40
solve this

I substituted y into it so I had:
2y*2-8y+16 = 40
2y*2-8y-24 = 0

I factorised and got:
(2y-12)(y+4)
Yep that's correct
(Original post by lemo****69)
Thanks!
You're welcome
11. (Original post by JLegion)
(2y-12)(y+4) = 0

So you know for that to be true, one (or both) of those brackets must equal zero.

2y -12 = 0
Y + 4 = 0

Can you take it from here?
Yeah, realised it had to be equal to 0 so I'd substitute 6 and -2 into the original equation. Thanks for the help!

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