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

Discussion in 'Community Discussion' started by sjjordan, Nov 8, 2006.

1. ### sjjordan macrumors 6502

Joined:
Jun 10, 2003
Location:
United States
#1
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

2. ### bearbo macrumors 68000

Joined:
Jul 20, 2006
#2
4*0 = 3*0 does not mean 4 = 3

3. ### Kolind macrumors regular

Joined:
Nov 2, 2006
Location:
Denmark
#3
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.

4. ### bearbo macrumors 68000

Joined:
Jul 20, 2006
#4
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...

5. ### atszyman macrumors 68020

Joined:
Sep 16, 2003
Location:
The Dallas 'burbs
#5
a+b = c

c=c

c-c = 0

a+b-c = 0

4*0=3*0

does not mean 4=3

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.

6. ### Kolind macrumors regular

Joined:
Nov 2, 2006
Location:
Denmark
#6
Looks like I was a bit too fast there In this case a+b-c=0 - not sure what I was thinking I second whatever Bearbo said... and will stop trying to write anything about math in English 7. ### MongoTheGeek macrumors 68040

Joined:
Sep 13, 2003
Location:
Its not so much where you are as when you are.
#7
Most times when you get problem like this the answer is you can't divide by 0.

8. ### bemylover macrumors regular

Joined:
Jun 20, 2005
#8
That's probably "always", rather than "most times".

9. ### fradac macrumors regular

Joined:
Oct 24, 2003
Location:
Atlanta, GA
#9
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

10. ### Rodimus Prime macrumors G4

Joined:
Oct 9, 2006
#10
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.

11. ### sassyk431 macrumors newbie

Joined:
Mar 10, 2014
#11
(4+3a)=9(3a)

Can some one explain how to do this problem?

(4+3a)=9(3a)

12. ### BenTrovato macrumors 68030

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Jun 29, 2012
Location:
#12
-4 - 3a = 27 - 9a
6a = 31
a = 31/6

13. ### ideal.dreams macrumors 68020

Joined:
Jul 19, 2010
Location:
Ohio
#13
You joined a MACRumors website and dug up a thread that is 8 years old for help with your math homework. Cool Distribute. -4-3a=27-9a
Combine like terms. 6a=31
Isolate variable. a=31/6

14. ### The Doctor11 macrumors 603

Joined:
Dec 15, 2013
Location:
New York
#14
I can get homework help on here? This site just got better 