I'm in a bit of jam. I need to solve the equation, abc=a for "a."?

Question

It appears to me that solving for "a" is not possible. Is this a so-called "conditional equation"?

Thanks everyone for the responses.

Answer 1

abc=a
if and only if
a=1, b=1 & c=1

In that case abc=a=b=c

Answer 2

Yes, it is conditional.

Here the conditions are
bc=1
or
a=0

(or = any one condition is sufficient)

if bc = 1 then we have abc = a(bc) = a*1 = a
This implies, of course that b and c are 'units': there exists values b and c such that b = 1/c and vice versa (with integers, this is only possible for b=c=1 and b=c=-1)

if a=0, then we have abc = a(bc) = 0*(bc) = 0 = a

If you are in real numbers, I cannot see any other condition that could work.

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PS:

both conditions can exist simultaneously (i.e., there is nothing forbidding a=0 AND bc=1)

Let a=0 and bc=1
abc = a(bc) = 0*1 = 0 = a

Answer 3

abc - a = 0
a(bc-1) = 0
therefore a = 0 OR bc-1 = 0 --> bc = 1

So you have an infinite number of solutions to the equation since A can equal 0 or A can be any other number as long as bc = 1.

P.S. Do not divide by A. Dividing by a variable is illegal, UNLESS you know it's not equal to zero.

Answer 4

Subtract 'a' from both sides:

abc - a = 0

factor:

a(bc - 1) = 0

So, either a = 0 or bc = 1.

When bc = 1, it doesn't matter what 'a' is.

Answer 5

abc = a

so "a" is 0

0*b*c = 0
(no matter what numbers b and c are when time by 0 it will be 0)

Hope it helps.

Answer 6

The only way i can see this being possible is if bc=1. thats the only way that "a" can hold its value in a multiplication problem.

Answer 7

divide both sides of the equation by a, giving
bc=1
If you put bc=1 back into the original equation, you have a(1)=a.

I would say a could be any number.

Answer 8

The problem is you are using jam!

a+b+c=a

Jelly+Peanut butter+ Bread (squared)=PB&J

And you wonder why I had to retake algebra (or is that geometry?)

Answer 9

a can be anything if and only if b*c = 1 (i.e. b and c are inverses, b = 1/c)

Answer 10

a=0 no matter what u multiply the answer is 0
(0)(2)(4)=(0)
(a)(b)(c)=(a)