Question

2/9j+2/3

factor out the coefficient of the variable :)

Answer 1

whats the question?

Answer 2

That is the question.

Answer 3

\[\large\rm \frac{2}{9}j+\frac{2}{3}\]Hey Maddie :)
Taking the entire 2/9 out of each term?
Hmm ok this will be a little tricky!\[\large\rm \frac{2}{9}\left(j+?\right)\]When we divide 2/9 out of 2/9j, it just leaves us with the j, ya?
The other one is where we really have to focus.

Answer 4

The other term is this,\[\large\rm \frac{\left(\frac{2}{3}\right)}{\left(\frac{2}{9}\right)}\]It's the 2/3 with the 2/9 divided out of it.
Remember how to divide fractions? :)

Answer 5

If that's too confusing, maybe we can try it a different way.

Answer 6

Yeah that's very confusing!

Answer 7

This way might also confuse you, but we can try at least.\[\large\rm \frac{2}{9}j+\frac{2}{3}\]Factoring out a 2 from each term gives us\[\large\rm 2\left(\frac{1}{9}j+\frac{1}{3}\right)\]And then we need to pull out the 1/9.
How many 1/9's can fit into 1/3? :)

Hmm ya this way sucks also... thinking...

Answer 8

Oh ok ok ok I got it.

Answer 9

If you think back to `adding fractions`, we always have to get a common denominator first, ya?

That's also going to be helpful here as well, so let's try that.
Hmm, looks our common denominator will be a 9.
So the second fraction needs a 3 multiplying the top and bottom.

Answer 10

ok im just confused. Sorry

Answer 11

ok

Answer 12

So we'll give this second fraction a 3 on top and bottom,\[\large\rm \frac{2}{9}j+\frac{2}{3}\color{royalblue}{\cdot\frac{3}{3}}\]We multiply straight across, so it becomes,\[\large\rm \frac{2}{9}j+\frac{6}{9}\]

Answer 13

ok so would it be 8/9j?

Answer 14

for the answer

Answer 15

Unfortunately, no.
The second number does not have a j connected to it,
so these are not `like-terms`.
They cannot be combined in any meaningful way.

Let's proceed by rewrite the 6 as 2*3.\[\large\rm \frac{2}{9}j+\frac{2\cdot3}{9}\]This next step might be a little tricky.
When you have a number in the numerator of a fraction, it can come off of the fraction as remain as multiplication next to the fraction.
We're going to take the 3 out of the numerator.\[\large\rm \frac{2}{9}j+\frac{2}{9}\cdot3\]

Answer 16

and* remain, typo

Answer 17

ok

Answer 18

I understand that the steps up to this point might be a little confusing,
but from this point, is it easier to see what we're left with when we take a 2/9 out of each term?\[\large\rm \color{royalblue}{\frac{2}{9}}j+\color{royalblue}{\frac{2}{9}}\cdot3\]

Answer 19

we would be left with jx3 (

Answer 20

x is a multiplication sign

Answer 21

No, not j*3.

Ignore the dot next to the 3 if that's confusing you.
This is our expression right now,\[\large\rm \color{royalblue}{\frac{2}{9}}j+\color{royalblue}{\frac{2}{9}}3\quad= \color{royalblue}{\frac{2}{9}}\left(~~?~+~?~~\right)\]We have addition, so we should still end up with addition in the end.
Just take the blue thing out of each term, what are you left with?

Answer 22

so j+3?

Answer 23

\[\large\rm \color{royalblue}{\frac{2}{9}}j+\color{royalblue}{\frac{2}{9}}3\quad= \color{royalblue}{\frac{2}{9}}\left(~j~+~3~\right)\]J+3 is what's left in the brackets, good.
And that is our final answer! :)

Maddie Maddie Maddie.
Madz.... you gotta practice the math more ;) Keep studying girl!!!

Answer 24

haha i will! thank you so much!!!!!

Answer 25

np