inequality in interval notation: ab - 5 = b/3

You've posted several very similar problems. It's time to start working the other way. Let's see what you get and how you get it. We'll be happy to review your well-written, easily-follow methodology.
 
mikey13 said:
ab - 5 = b/3

solve for b
Click to expand...
Without a value for "a", this cannot be solved for a numerical value of "b". But you can solve symbolically.

* Get the terms containing "b" together on one side, with any other terms on the other side.

* Factor out the "b".

* Divide off whatever you factored the "b" from.

* Simplify, as necessary.

If you get stuck, please reply showing how far you have gotten. Thank you.

Eliz.
 
well this is what i got

ab-5=b/3
(3)ab-5=3(b/3)
3ab-3ab-5=b-3ab
-5=b(1-3a)
-5/1-3a=b


can anyone tell me if its right? Still cant figure out how to put it on a number line :?
 
Re: well this is what i got

ab-5=b/3
(3)ab-5=3(b/3)

Right off. Very bad notation. Should be (3)[ab-5]=3(b/3)

3ab-3ab-5=b-3ab

See. You confused yourself. Should be "-15". I have no idea where you managed two copies of "ab". Why did you do that?

-5=b(1-3a)

You are solving for 'b'. You just put 'a' over there with it. Don't do that.

-5/1-3a=b

Again with the bad notation. Use parentheses liberally to clarify meaning.

-5/(1-3a) = b

Start it again. No errors, this time.
 
OK, you copied my last line. Now go back and start over. This time, fix all the other errors I pointed out.
 
inequality in interval notation

ab-5=b/3
(3)[ab-5]=3(b/3)
3ab-15=b
3ab-3ab-15=b-3ab
-15=b(1-3a)
-15/1-3a=b(1-3a)/1-3a
-15/1-3a=b
 
Re: inequality in interval notation

mikey13 said:
ab-5=b/3
(3)[ab-5]=3(b/3)
3ab-15=b
3ab-3ab-15=b-3ab
-15=b(1-3a)
-15/1-3a=b(1-3a)/1-3a
-15/1-3a=b
Click to expand...
So Close!! The last two steps need some help.

-15/(1-3a)=b(1-3a)/(1-3a)
-15/(1-3a)=b

That's better.

Note: I am NOT just being picky. It was wrong the way it was. Think about the Order of Operations and you will see it.
 
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