anonymous
  • anonymous
2/9j+2/3 factor out the coefficient of the variable :)
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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nathalie0918
  • nathalie0918
whats the question?
anonymous
  • anonymous
That is the question.
zepdrix
  • zepdrix
\[\large\rm \frac{2}{9}j+\frac{2}{3}\]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.

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zepdrix
  • zepdrix
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? :)
zepdrix
  • zepdrix
If that's too confusing, maybe we can try it a different way.
anonymous
  • anonymous
Yeah that's very confusing!
zepdrix
  • zepdrix
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...
zepdrix
  • zepdrix
Oh ok ok ok I got it.
zepdrix
  • zepdrix
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.
anonymous
  • anonymous
ok im just confused. Sorry
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
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,\[\large\rm \frac{2}{9}j+\frac{6}{9}\]
anonymous
  • anonymous
ok so would it be 8/9j?
anonymous
  • anonymous
for the answer
zepdrix
  • zepdrix
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.\[\large\rm \frac{2}{9}j+\frac{2}{9}\cdot3\]
zepdrix
  • zepdrix
and* remain, typo
anonymous
  • anonymous
ok
zepdrix
  • zepdrix
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?\[\large\rm \color{royalblue}{\frac{2}{9}}j+\color{royalblue}{\frac{2}{9}}\cdot3\]
anonymous
  • anonymous
we would be left with jx3 (
anonymous
  • anonymous
x is a multiplication sign
zepdrix
  • zepdrix
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?
anonymous
  • anonymous
so j+3?
zepdrix
  • zepdrix
\[\large\rm \color{royalblue}{\frac{2}{9}}j+\color{royalblue}{\frac{2}{9}}3\quad= \color{royalblue}{\frac{2}{9}}\left(~j~+~3~\right)\]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!!!
anonymous
  • anonymous
haha i will! thank you so much!!!!!
zepdrix
  • zepdrix
np

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