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Yeah but that's doesn't really help where I'm going with it though. An example of what I'd be kinda trying to get to would be:
3x+2y+5
3z+14
And then just doing:
3x+2y+5 / 3z
and
3x+2y+5 / 14
You see? What I'm trying to do isn't in the book or anything, I was just curious why it wouldn't work.
It doesn't work because that's not how division works. I'm not sure of the best way to explain it, but basically the easiest way to understand why this won't work is to ignore the idea of "division," and instead thinking of division as multiplication by the inverse (which it is). So instead of dividing by 6, you multiply by 1/6.
Then just realize that 1/(x+y) does not equal 1/x + 1/y...it equals A/x + B/y, where A and B are values to be determined though a fun process known as partial fraction decomposition.
If you're in need of proof that 1/(x+y) != (1/x)+(1/y), then just substitute constants in for x and y to prove it to yourself.
To be more specific, multiplication has the distributive property, which basically states that a * (b + c) = a * b + a * c. So, as above, 10 * (4 + 2) = 10 * 4 + 10 * 2.
Division doesn't have that property, as stated well in the post above
Yeah but that's doesn't really help where I'm going with it though. An example of what I'd be kinda trying to get to would be:
3x+2y+5
3z+14
And then just doing:
3x+2y+5 / 3z
and
3x+2y+5 / 14
You see? What I'm trying to do isn't in the book or anything, I was just curious why it wouldn't work.
It doesn't work because that's not how division works. I'm not sure of the best way to explain it, but basically the easiest way to understand why this won't work is to ignore the idea of "division," and instead thinking of division as multiplication by the inverse (which it is). So instead of dividing by 6, you multiply by 1/6.
Then just realize that 1/(x+y) does not equal 1/x + 1/y...it equals A/x + B/y, where A and B are values to be determined though a fun process known as partial fraction decomposition.
If you're in need of proof that 1/(x+y) != (1/x)+(1/y), then just substitute constants in for x and y to prove it to yourself.
To be more specific, multiplication has the distributive property, which basically states that a * (b + c) = a * b + a * c. So, as above, 10 * (4 + 2) = 10 * 4 + 10 * 2.
Division doesn't have that property, as stated well in the post above
Thanks guys, I understand it a lot better now, it was just one of those things like, 'hmm, it SHOULD work, since it works with multiplication, and division is just multiplying by the recip' but for some reason my mind just wasn't working.
Posts
60 / 10 = 6
40 / 10 = 4
20 / 10 = 2
3x+2y+5
3z+14
And then just doing:
3x+2y+5 / 3z
and
3x+2y+5 / 14
You see? What I'm trying to do isn't in the book or anything, I was just curious why it wouldn't work.
so for your example the appropriate thing to do would be
60 * 1/20
When you do 60 ÷ 5 you're multiplying the divisor (10 or 1/10) by 2 instead of dividing it by 2.
(3x/3z+14)+(2y/3z+14)+(5/3z+14)
To be more specific, multiplication has the distributive property, which basically states that a * (b + c) = a * b + a * c. So, as above, 10 * (4 + 2) = 10 * 4 + 10 * 2.
Division doesn't have that property, as stated well in the post above
Division has distributive property. Example: (4+2)/2 = (4/2)+(2/2) = 2+1 = 3
If you have
3x+2y+5
3z+14
you can break it down, but it's 3x / (3z+14) + 2y / (3z+14) + 5 / (3z+14)
what you need to realize is you're dividing by a quantity 3z+14...and based on the order of ops the parenthesis come before the division
Thanks for the help.