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(a-b)/a = c
I need to find a =
On University Challenge it's either 1, -1 or 0 as the answer to any maths question, have you tried those?
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Curse you, Mastiles!
Seriously I think it is 1.
But I always was shocking at algebra so it might well be twelvety seven.
No it is not. It can't be.
I think.
Sorry I meant I need to find a in terms of b and c.
(a-b)/a = c
a-b = ca
a-ca = b
a = b/(1-c)
A=(CA)+B
a = (c-1)/-b
(a-b)/a = c
a-b = ac
a-ac = b
a(1-c) = b
a = b/(1-c)
Ohh. Interesting! 🙂
What IS the point of algebra anyway? I never did understand it at all.
🙁
wot off the brakes said
As a complete aside (as thepurist seems to have solved it)...
Friday nights in the local we've taken to doing the landlady's 11yo son's maths homework. Not for him but so she knows whether he's got it right. Now I've got an A-level in Maths and a degree (Signal Processing) that had a fair bit of maths in it but there's some weeks we really struggle. Just thought you'd like to know.
Being as I'm wrong, can someone please walk me through the solution? 🙂
Thanks chaps. Those GCSE maths lessons are a very distant memory.
Being as I'm wrong, can someone please walk me through the solution?
(a-b)/a = c
a-b = ac (divide both sides by a)
a-ac = b (swap ac and b from one side to the other, this changes their signs)
a(1-c) = b (extract a common factor of a on the left)
a = b/(1-c) (divide both sides by (1-c))
Being as I'm wrong, can someone please walk me through the solution?
What offthebrakes said.
Thanks Feenster, been a LONG time since I did anthing so pointless as algrbra ( 😉 ) I'd forgotten how it all works 🙂
[url= http://www.wolframalpha.com/input/?i=%28a-b%29%2Fa+%3D+c ]Wolfram|Alpha is your friend[/url]
feenster - Member(a-b)/a = c
a-b = ac (divide both sides by a)
a-ac = b (swap ac and b from one side to the other, this changes their signs)
a(1-c) = b (extract a common factor of a on the left)
a = b/(1-c) (divide both sides by (1-c))
I'm being thick. Can you explain how a-ac becomes a(1-c)
I can't believe I used to be good at algebra 🙁
a-ac is the same as (a x 1) - (a x c), so you can take the common factor a out and make it a x (1-c).
Anyone for Sturm Liouville systems?
... for non-zero a, and c not equal to 1.
(a-b)/a=c
a-b= ac
a-ac=b
a(1-c)=b
a=b/(1-c) as long as neither a=0 nor c=1