Four equals to three

If a-b=c then it is obvious that a-b-c=0. For example x-3y=6 implies that x-3y-6=o

The starting equation states that ----

$a-b=c \implies a-b-c=0$ .

Now, when Kimmie cancelled both the sides in her work, what she actually did was dividing both sides of the equation $4(a-b-c)=3(a-b-c)$ by $(a-b-c)$ so as to get $4=3$ but since $(a-b-c)=0$ , so she actually divided both sides of the equation by $0$ which is mathematically incorrect.

So, her main error in the work was Dividing both sides by 0.

The answer is obvious. Division by 0 is strictly prohibited, which causes no solution already. No solution means that 4 != 3.

Since from the original equation $a - b = c$ , then $a - b - c = 0$ . In the factoring, she divided by $a - b - c = 0$ , and thus, that is the answer.

lets start with -

a - b = c

a - b - c = 0

4 ( a - b - c ) = 4 ( 0 ) = 0 --- eqn (1)

3 ( a - b - c ) = 3 ( 0 ) = 0 ---eqn (2)

dividing eqn. 2 by eqn 1 ,

4 ( a- b - c ) / 3 ( a - b - c ) = 0 / 0

Here , Kimmie took 0 / 0 to be equal to 1, thereby getting 4 = 3 as the result.

However , as we know, 0 / 0 is not defined , this was a wrong step. Hence the answer.

a-b=c so, a-b-c=0

4(a-b-c)=3(a-b-c) because a-b-c = 0, then 4 (a-b-c) = 3 (a-b-c) should not be divided by (a-b-c) or in other words should not be divided by 0.

Okay jump to the last equation

4(a - b - c) = 3(a - b - c)

4(a - b - c) - 3(a - b - c) = 0

(a - b - c)( 4- 3 ) = 0

Therefore a - b - c has to be = 0

* And therefore in the last step we cannot cancel by a - b - c as it would be equivalent to dividing by 0 *

$a-b=c$ $a-b-c=0$ Since you cannot divide by 0, the conclusion is proven.

By the problem if 4(a-b-c)=3(a-b-c). But earlier she started with a-b=c; or this can also be written as a-b-c=0. Therefore 4(a-b-c)=3(a-b-c), which can also be written as 4(0)=3(0). Hence she divided by 0 is the correct answer.

Aman Jaiswal is right. Obviously, $a-b-c=0$ .

a major mistake, a-b=c then a-b-c=0 , and any constant divided by 0 equals undefined (or infinity)

Kimmie took a-b=c, so a-b-c=0. At the final stage She wrote 4(a-b-c)=3(a-b-c), that is 4 0=3 0 which gives 0=0, she can't divide both sides by (a-b-c) for cancellation, as it is 0. It was the main error that she divided by 0.

We have a - b = c <=> a - b - c = 0. Kimmie cannot both sides 4(a - b - c) = 3(a - b - c).

when the equation turned to, 4(a-b-c)=3(a-b-c) we look back to the guiding equation, a-b=c, which yields a-b-c=0 which means we are dividing by zero.

as a-b=c the a-b-c=0 so it becomes 0/0 which is incorrect

its called divide by zero exception in algebra, step 1 says [a-b=c that is a= b+c ] and second last step 4[a-b-c]=3[a-b-c] replacing a-b with c , we get 4[c-c]=3[c-c] that is 4x0=3x0 , last step we "divide by zero" as 4x0/0= 3x0/0 which is not legit.

If we solve this the usual algebraic way, we won't find anything unusual. But this Mathematical fallacy obviously calls for a little more careful analysis.

If you see the first step itself ( a - b = c ) and shift c to the left side, it becomes a - b - c = 0

So practically in the final step, you are dividing by 0! And that also without realizing because we always try to eliminate as many unknowns as possible.

a-b-c=0 because a-b=c, so moving c to the left makes a-b-c=0.

she assumed a-b=c →a-b-c=0 in the final step she cancelled (a-b-c) from both sides. but (a-b-c)=0

a - b = c

a - b - c = 0

As we can see that a-b=c , therefore a-b-c=0 and we cannot cancel the terms on both sides, which results to zero.

She assumed that (a-b=c) which implies (a-b-c=0) , later she divided( a-b-c ) on both the sides and there lies the error.

she did the mistake at first line. because.

a - b = c

and,

a - b - c =0

when she try to clear both side with (a - b - c) , actually she divided with 0. and 0/0 is no valid.

Aaa... Ok, I got it.... When she factored it out, the statement 4(a-b-c)=3(a-b-c) became 4(0)=3(0) then she divided it both sides by zero...

a-b=c hence a-b-c=0 therefore mistake is that it was divided by zero

Now this one was pretty simple, folks. The first assumption itself says : a-b=c, which means that a-b-c=0 . Whereas in the last step, the same term: a-b-c is cancelled out, but we cant cancel out zero. So, that is the error!

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

a-b-c = 0

a-b-c=0 so divide by 0 got it error

you cannot cancel a-b-c as that is zero

4th step is incorrect because from 1st step; a-b=c;which implies a-b-c=0 and the error in the first step is we are dividing a-b-c/a-b-c=0/0 which doesnot exist

if a-b = c a-b-c = 0 so, it cannot be cancelled out.

Given: (a-b)=c. Replace (a-b) with c before cancelling out in last step. We get 4(0)=3(0). :P

a - b = c so a - b - c = c - c = 0

Nothing can be divided by 0. So it is the main error of Kimmie,s.

The starting equation states that ----

.

Now, when Kimmie cancelled both the sides in her work, what she actually did was dividing both sides of the equation by so as to get but since , so she actually divided both sides of the equation by which is mathematically incorrect.

So, her main error in the work was Dividing both sides by 0.

We start with a-b = c;

This means a-b-c = 0.

In last step both sides are divided by a-b-c, which is zero division and in valid

since a-b=c implies a-b-c=0 how can she divide an equation with zero?? thats the error....

a-b=c

a-b-c=0

4(a-b-c)=3(a-b-c)

4(0)=3(0)

0=0

She devided by 0

it's easy, as a-b=c; thus a-b-c=0 and if both the sides of an equation carry 0, we cannot change their sides as by changing sides the finite equation will become infinite and incorrect.

a-b=c so a-b-c=0

IN THE 4th STEP SHE WROTE : 4(a-b-c) = 3(a-b-c) NOW , SHE DIVIDED BOTH THE SIDES BY (a-b-c). BUT SHE ASSUMED THAT a-b =c => a-b-c = 0 THEREFORE SHE DIVIDED BOTH THE SIDES BY 0.

After the factorization 4(a-b-c) = 4((a-b)-c) = 4(c-c) since a-b=c. as such the L.H.S. = 4x0 and R.H.S is 3x0. 4=3 can only be obtained if there is a division by zero.

She started off with the equation a-b=c =>a-b-c=0

but at the end she got 4(a-b-c)=3(a-b-c) =>4x0=3x0 =>0=0

But before that she cancelled a-b-c from both side which is impossible. Because for this she need to divide by 0 in both sides, which is impossible. So this is wrong.

she shouldnt have cancelled in the last step!!

Since a-b=c, Then (a-b-c) = 0

Here a-b=c,so,a-b-c=0,so.we can not cancel both side (a-b-c) in last step.

To cancel any variable term in an algebraic eqation, it is necessary to make sure that the term which is being cancelled is not zero...as dividing on both the sides by zero is INDETERMINATE form..

a-b=c a-b-c=0 so at last step we cannot cancel (a-b-c) terms because its indeterminate number '0' (i.e.) its logical error