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4=3 Math Problem. Can you figure it out?

S

sjjordan

macrumors 6502
Original poster
Jun 10, 2003
296
1
United States
I ran into this math problem earlier today and it took me a while, but I figured out what's wrong with it. Thought I'd pass it on for fun. Enjoy!

Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganising:
4a + 4b - 4c = 3a + 3b - 3c
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
Remove the same term left and right:
4 = 3
 
sjjordan said:
I ran into this math problem earlier today and it took me a while, but I figured out what's wrong with it. Thought I'd pass it on for fun. Enjoy!

Suppose:
a + b = c
This can also be written as:
4a - 3a + 4b - 3b = 4c - 3c
After reorganising:
4a + 4b - 4c = 3a + 3b - 3c
Take the constants out of the brackets:
4 * (a+b-c) = 3 * (a+b-c)
Remove the same term left and right:
4 = 3
Click to expand...

4*0 = 3*0 does not mean 4 = 3
 
Kolind said:
a+b-c does not necessarily give zero... the problem is that you can't remove the same term on both sides as it's being multiplied with the two constants.
Click to expand...

a+b=c therefore a+b-c = 0

you most certainly can remove the same term from both side if they're being multiplied both side, by constant or variable... but you cannot remove term if there's , for instant, an addition...

i.e. a*b+c = e*b+f, you cannot then remove the b to make it a+c = e+f...
 
Kolind said:
a+b-c does not necessarily give zero... the problem is that you can't remove the same term on both sides as it's being multiplied with the two constants.
Click to expand...

a+b = c

c=c

c-c = 0

a+b-c = 0

4*0=3*0

does not mean 4=3

bearbo said:
a+b=c therefore a+b-c = 0

you most certainly can remove the same term from both side if they're being multiplied both side, by constant or variable... but you cannot remove term if there's , for instant, an addition...

i.e. a*b+c = e*b+f, you cannot then remove the b to make it a+c = e+f...
Click to expand...

the removal of the term is legitimate, sort of.

for x*y = z*y you can divide both sides by y and determine x=z.

However if y=0 you cannot do this since you are dividing by zero which is undefined....

I guess you're both right.

if a+b were not equal to c then you could remove the (a+b-c) by dividing both sides by (a+b-c). But when you do this the following equation is no longer true:

4a - 3a + 4b - 3b = 4c - 3c

That equation is only true when a+b = c which means (a+b-c) = 0 and you cannot factor it out because it requires dividing by zero.
 
Looks like I was a bit too fast there ;)

In this case a+b-c=0 - not sure what I was thinking :confused:

I second whatever Bearbo said... and will stop trying to write anything about math in English :p
 
yup its another one of those "cant cancel 0 on both side of equation"

here is another example

a = b = 1

a*a = a*b

a*a - b*b = a*b - b*b

(a+b)(a-b) = b(a-b)

a+b = b

1+1 = 1

2 = 1
 
umm again you have a bunch of zeros in there.

a=b=1
a-b=0
So most of you work falls apart when almost all the proof = zero. You just removed. oh and 0/0 is undefined so it falls apart there as well

I going to show you exactly where it fails.
(a+b)(a-b)=b(a-b)
You divided out the a-b. Only one problem a-b=0 and and 0/0=undefined. that makes the next step invalided.
 
sassyk431 said:
Can some one explain how to do this problem?

–(4+3a)=9(3–a)
Click to expand...

You joined a MACRumors website and dug up a thread that is 8 years old for help with your math homework. Cool :cool:

Distribute. -4-3a=27-9a
Combine like terms. 6a=31
Isolate variable. a=31/6
 
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