[SPLIT] solving s = L + 2b for b; I did s = L +2b/2b = L + 1

[GbbxO]

I have this question im trying to solve.

-> s= l+2b ; b

i wrote it out like this,

s= l+2b/2b
cancelled out 2b

s/(l)= l/(l)

So my solution is..

s/l= b <- is that right ??

 

[Denis]

What grade are you in, Gb?
You are evidently completely unaware of how to solve basic equations...only your teacher can help you.

 

[mmm4444bot]

s = l + 2b ; solve for b

s= l+2b/2b
cancelled out 2b

Hello there:

Here are two comments about your attempt to divide by 2b.

Firstly, algebra does not allow you to divide just one part (or just one side) of an equation.

If you want to divide by 2b, then you must divide both sides of the equation by 2b. It would go like this:

$\displaystyle \frac{s}{2b} \;=\; \frac{l + 2b}{2b}$

$\displaystyle \frac{s}{2b} \;=\; \frac{l}{2b} + \frac{2b}{2b}$

$\displaystyle \frac{s}{2b} \;=\; \frac{l}{2b} + 1$

You could begin this way, but you're making extra work for yourself because now the symbol for which you want to solve is in the denominator of a fraction. You would eventually need to multiply both sides by b, so that kinda defeats the purpose of dividing in the first place.

Secondly, after you illegally divided only a portion of the equation, you ended up with a new (wrong) equation.

s = l

An alarm bell should have gone off at this point.

Do you realize what your goal is? It is to solve for the symbol b. This means that your final result must be in the form of b = something.

Where is b in your result s = l ? There is no b. (The b be gone.) You had no reason to continue, at that point, since the symbol you're trying to isolate vanished.

Start over from the beginning by getting the term 2b all by itself on one side of the equal sign. Can you see what you need to subtract from both sides in order to get 2b all by itself? After doing this subtraction, you would next divide both sides by the coefficent of b (that is 2) in order to get an equation of the form b = something.

Then, you're done.

Thank you for showing your work.

Cheers,

~ Mark