Question

Select the appropriate restrictions for the variables.
6a+12 / a+2


A.
a≠–2

B.
a≠2

C.
a≠6

D.
There are no restrictions on the values of variable a.

Answer 1

So, I'm guessing the problem is like this:
We want \(\dfrac{6a+12 }{ a+2}\) to be an existing number.

But denominators can't be \(0\), because just try and divide something by \(0\). It makes no sense. So it's something you can write down, but it's meaningless mathematically.

So we need to make sure that the denominator isn't \(0\). That variable \(a\) can ruin it for us. It can make the denominator be \(0\).

Buuuuuuut, if we "restrict" it, so it can be some numbers and not others, then we can say that your fraction there is okay. Like, if \(a\) is going to make the denominator zero, we simply say, "hey, \(a\), you can't be that value that makes the denominator \(0\)!".

Answer 2

So do you have any thoughts on this problem?

Answer 3

Um...not really, I'm still kinda confused :/ Am I supposed to find the number that's a? or what a represents?

Answer 4

Well, what you need to do, getting straight to the point, is find what \(a\) can be to make the denominator \(0\). Then you want to restrict \(a\) by saying it cannot be that value that makes the denominator zero.

So we look at \(\dfrac{6a+12 }{ a+2}\)

The denominator is then \(a+2\), right?

Answer 5

ya

Answer 6

Okay, so what \(a\) will make it \(0\)?

Here's a good way to do it.
Suppose that the denominator \(\sf does\) equal \(0\), and then solve for \(a\).
So,
\(a+2=0\implies a=?\)

Answer 7

2? Because can't you replace the 0 with 2? :/

Answer 8

Are you solving for \(a\)?

Answer 9

I dont really know....I'm kinda confused on how to solve for a, am I supposed to use the top numbers?

Answer 10

Nope! Just \(a+2=0\).
Are you taking algebra?

Answer 11

Yeah

Answer 12

Okay! So, you know that you want to get \(a\) to its own side of the equation.

But you see that you're adding \(2\) to it. How do you get rid of a \(2\) that's being added?

Answer 13

um..You can use the 6 and 12 I think..sorry I'm really confused :/

Answer 14

Well, forget about the entire problem.

....

Now, think about this problem:
\(a+2=0\)
Find \(a\).

Answer 15

Can you find the answer to that problem?

Answer 16

Um well, If you can use negatives then I kinda think it would be -2..

Answer 17

Looks good!
\(\ \ \ a+2=0\)
\(\ \downarrow\ \ \ \ \downarrow\ \ \ \downarrow\)
\(-2+2=0\)

Answer 18

so that's right..?

Answer 19

So, think about this.
When \(\color{blue}{a=-2}\), then \(\color{red}{a+2=0}\).

So, when you have
\(\dfrac{6a+12 }{\color{red}{ a+2}}\)
and \(\color{blue}{a=-2}\)
then
\(\dfrac{6a+12 }{\color{red}{ \color{blue}{-2}+2}}\)
\(=\dfrac{6a+12 }{0}\) which is really not a number.

So we can't have \(\color{blue}{a=-2}\). So we \(\sf restrict\) \(a\) to not be \(2\).

So the answer is
\(a\neq2\) (is our restriction)

Answer 20

ohh ok, so I could have just said 2? instead of -2?

Answer 21

You have to say that \(a\) CANNOT be \(2\).

So you say it like \(a\) does not equal \(2\) by writing \(a\neq2\)

Answer 22

I mean negative 2!

Answer 23

So, correction:

You have to say that \(a\) CANNOT be \(-2\).

So you say it like \(a\) does not equal \(-2\) by writing \(a\neq-2\)

Answer 24

ohh ok Thank you! :D

Answer 25

If \(a\) does equal \(2\), then things are bad because then
\(\quad\dfrac{6a+12 }{ a+2}\)
is not a number anymore. It's "undefined." It makes no sense!

You're welcome!

Answer 26

Thank you :)

Answer 27

You're welcome! :)