in math, does ab+c= b+ac?

ab+c = b+ac

clearly not ! As you said ; you had a bit of a brain block!

whether you call rewriting 5b+40 with brackets as a simplification is for you to decide

but 5b+40 is the same as adding 8 to b then multiplying the answer by 5 , written algebraically as

5(b+8)

Only for specific values of a, b, and c... the real answer to your question is no, but there exist values for a, b, and c (a specific single set of values) that would make it true.

To find these values, we simply solve for one of the variables.

As a simple example, if a, b, and c all equal 1:

1*1 + 1 = 1 + 1*1

1 + 1 = 1 + 1

2 = 2

true

So this is true if a = b = c, because in this case, it doesn't matter what a, b, or c are as long as they are equal:

let a = x, let b = x, let c = x

x*x + x = x + x*x

x^2 + x = x + x^2

x(x + 1) = x(1+x)

x + 1 = 1 + x

this is true for any value x

If, however, a, b, and c are not all the same, there are still certain specific values for a, b, and c that would make this equation true. For example, let a = 1, and b and c could be any values:

ab + c = b + ac

1b + c = b + 1c

b + c = b + c

However, for most values of a, b, and c, the equation does not hold true. To illustrate, let's pick 2, 3, and 4:

2*3 + 4 = 3 + 2*4

6 + 4 = 3 + 8

10 = 11

not true

Hope this helps!

Edit: I don't usually like to contradict others, but I must point out that Jeff's declaration that "The original equation is only true if a = 1 or b = c, otherwise it's false." is not accurate.

Let a = 2, b = 4, and c = 6:

2*4 + 6 = 4 + 2*6

8 + 6 = 4 + 12

14 = 14

ab + c =? b + ac

Sometimes, but the answer is no.

5b + 40 = b + 200

4b = 160

b = 40

In this case, when a = 5, b = 40 and c = 40 the answer is yes. But this is just because b and c are equal. If b and c are equal, you can rewrite the equation as:

ab + b = b + ab

b(a + 1) = b(1 + a)

That is true.

Generally speaking, no. For instance, let's take b = 1; you're saying that 5 + 40 = 1 + 200...last I checked, 54 and 201 were very different numbers.

no

a = 2

b = 5

c = 40

ab + c = 10 + 40 = 50

b + ac = 5 + 80 = 85

ab + c = b + ac

ab - ac = b - c

a(b - c) = b - c

a = (b - c) / (b - c)

a = 1, unless b - c = 0

a = 1, unless b = c

The original equation is only true if a = 1 or b = c, otherwise it's false.

5b + 40 = b + 200

5b - b = 200 - 40

4b = 160

b = 160/40

b = 40

The original equation is only true if b = 40, otherwise it's false.

No.

Counterexample: a = -1, b = 2, c = 3

ab+c = 1

b+ac = -1

It holds for certain values of a,b,c, but not in general. In that sense, the equation is a constraint, not an identity.

it depends on the variable

clearly no .. just substitute any value for b and you'll see it is wrong