Simultaneous Equations help!
I'm doing simultaneous equations and I've got to (2y-12)(y+2), how do I work out y at this point?
Original post by lemoslag69
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.
Original post by lemoslag69
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!
Please Post the original question.
(edited 7 years ago)
Original post by lemoslag69
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
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)
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)
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
(2y-12)(y+2) = 0
2y = 12
y = 6
or
y = -2
So the answer is y = 6, y = -2
Thanks!
Original post by lemoslag69
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)
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?
Original post by lemoslag69
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)
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 lemoslag69
Thanks!
You're welcome
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?
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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