Algebra Manipulation Watch

H3rrW4rum
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#1
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Hey guys, what have I done wrong here?

If  \frac {a}{b} = 0 then (provided b is not infinity), a must equal 0.

However, what if I multiplied it all by c.

Would I get  \frac {a.c}{b} = 0

So this time  a.c = 0, a = 0 / c = 0 ?

I take it this is because you just multiplied by c, which you are now saying is 0, so you are saying you multiplied it by 0?

Also, are you even "allowed" to multiply both sides by c? As one side the c will "stay" but on the RHS the c will just multiply with the 0 and disappear

Thanks.

P.S I did this when I had the following question:

 \frac {1}{sinx} - \frac {4sin2x}{sinx} = 0

So instead of going  \frac {1 - 4sin2x}{sinx} = 0 \Rightarrow 1 = 4sin2x

I did  \frac {sinx - 4sin2xsinx}{sin^2x} = 0 \Rightarrow sinx (1 -4sin2x) = 0

So again by multiplying this time by sinx, I get an extra solution of sinx = 0.

So I am guessing whenever you have something equals 0, then multiply by something else, you get an extra solution where that something else is 0?
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Bobifier
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Why multiply by c? Just multiply by b, which will give you a = 0.

Also, with the original question, can you not just go from 1 = 4sin2x to sin(2x) = 1/4, solve for 2x and find x?
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H3rrW4rum
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(Original post by Bobifier)
Why multiply by c? Just multiply by b, which will give you a = 0.
I am not doing it for the right answer really: I am just wondering why you couldn't multiply it by something else.
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mathsisfunny
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I don't think so; I think when you factorize something it means that one or the other expressions must equal 0 not necessarily both.
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Pheylan
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(Original post by H3rrW4rum)
So again by multiplying this time by sinx, I get an extra solution of sinx = 0.
Yeah but sinx=0 isn't a valid solution
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Bobifier
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(Original post by H3rrW4rum)
I am not doing it for the right answer really: I am just wondering why you couldn't multiply it by something else.
You could happily multiply it by something else, and get exactly the result you suggested. There is mathematically no problem with that. The only problem is that I'm not absolutely sure you would want to, I don't think you actually gain anything from doing so.

Also yes, watch out for sinx = 0, as others have pointed out.
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H3rrW4rum
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(Original post by Pheylan)
Yeah but sinx=0 isn't a valid solution
Why not?

EDIT: Just tried a -b = 0

Let a and b = 5.

So multiply by c again, which I am calling 3,

ac - bc = 0 ( 3x5 - 3x5 = 0)

c( a - b) = 0,

3( 5 - 5) = 0

a - b = 0, but c =/= 0.

So the thing you multiply it by won't be 0, as the original part was already 0?
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Pheylan
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(Original post by H3rrW4rum)
Why not?
Because if you try to substitute it into the original equation, you get \dfrac{\mathrm{stuff}}{0}
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cooper21
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(Original post by H3rrW4rum)
Hey guys, what have I done wrong here?

If  \frac {a}{b} = 0 then (provided b is not infinity), a must equal 0.

However, what if I multiplied it all by c.

Would I get  \frac {a.c}{b} = 0

So this time  a.c = 0, a = 0 / c = 0 ?

I take it this is because you just multiplied by c, which you are now saying is 0, so you are saying you multiplied it by 0?

Also, are you even "allowed" to multiply both sides by c? As one side the c will "stay" but on the RHS the c will just multiply with the 0 and disappear

Thanks.
if you have a/b = 0 you can say a = 0
then if you want to multiply by c you have to do it both sides so you get ac/b = 0c
so basically c can be anything including 0
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