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# C1 maths help! watch

1. Hey dont know what i got wrong but when i substitute to check if the set of values are right it is wrong the the second equation but right for the first.
Simultaneous equations:
2u+v=7
uv=6

v=7-2u
u(7-2u)=6
7u-2u^2=6
-2u^2+7u-6=0
(u+4)(u+3)=0
u=-4 and u=-3

sub into eq 3: v=7-2(-4)
v=15

sub other value: v=7-2(-3)
v=13

Values are: v=15 u=-4
or v=13 u=-3

2. (Original post by YsfAli)
Hey dont know what i got wrong but when i substitute to check if the set of values are right it is wrong the the second equation but right for the first.
Simultaneous equations:
2u+v=7
uv=6

v=7-2u
u(7-2u)=6
7u-2u^2=6
-2u^2+7u-6=0
(u+4)(u+3)=0
u=-4 and u=-3

sub into eq 3: v=7-2(-4)
v=15

sub other value: v=7-2(-3)
v=13

Values are: v=15 u=-4
or v=13 u=-3

Pretty sure it was the error in factorizing that cause this mistake
3. (Original post by YsfAli)
Hey dont know what i got wrong but when i substitute to check if the set of values are right it is wrong the the second equation but right for the first.
Simultaneous equations:
2u+v=7
uv=6

v=7-2u
u(7-2u)=6
7u-2u^2=6
-2u^2+7u-6=0
(u+4)(u+3)=0
u=-4 and u=-3

sub into eq 3: v=7-2(-4)
v=15

sub other value: v=7-2(-3)
v=13

Values are: v=15 u=-4
or v=13 u=-3

Just an error factorising. I would move the 7u-2u^2 to the other side personally, just makes it easier having a positive term first. Then it factorises to (2u-3)(u-2)=0.
4. 2u^2-7u+6=0

becomes

2u^2-4u-3u+6=0

therefore

2u(u-2)-3(u-2)

which is same as saying

(2u-3)(u-2)

So..

u=3/2 or u=2
5. (Original post by mariaaax)
2u^2-7u+6=0

becomes

2u^2-4u-3u+6=0

therefore

2u(u-2)-3(u-2)

which is same as saying

(2u-3)(u-2)

So..

u=3/2 or u=2
Thank you so much i need to wake up !! lol
6. (Original post by YsfAli)
Thank you so much i need to wake up !! lol
Anytime

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Updated: September 9, 2015
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