I think this is the first question I've ever asked on this site in my whole time being on here :P
I'm in geometry right now, and I forget how to do this algebra!! I don't need an answer, just how to solve it :) (either one would be fine, though)
(x2 + 9x +14) • 9/3x + 21 (9 over 3x + 21, the whole thing)
Hi Saseflower,
It is WONDERFUL to see you!!
(x2 + 9x +14) • 9/3x + 21 (9 over 3x + 21, the whole thing)
I think the meaning of your question is clear but this is ideally how it could hve been presented.
(x^2+9x+14) * 9/(3x+21)
\((x^2+9x+14) * \frac{9}{3x+21}\\ =(x+2)(x+7) * \frac{9}{3(x+7)}\\ \mbox{ (x+7) cancels out because one is on the top and one is on the bottom.}\\ =(x+2) * \frac{9}{3}\\ =(x+2) * \frac{\not{9}^3}{\not{3}^1}\\ =(x+2) * 3\\ =3(x+2)\\ =3x+6 \)
I have not explained the initial factoring that I did in the first line. It is probably the hardest bit to understand.
Do you need me to explain more?? Which bits give you trouble.
Hi Saseflower,
It is WONDERFUL to see you!!
(x2 + 9x +14) • 9/3x + 21 (9 over 3x + 21, the whole thing)
I think the meaning of your question is clear but this is ideally how it could hve been presented.
(x^2+9x+14) * 9/(3x+21)
\((x^2+9x+14) * \frac{9}{3x+21}\\ =(x+2)(x+7) * \frac{9}{3(x+7)}\\ \mbox{ (x+7) cancels out because one is on the top and one is on the bottom.}\\ =(x+2) * \frac{9}{3}\\ =(x+2) * \frac{\not{9}^3}{\not{3}^1}\\ =(x+2) * 3\\ =3(x+2)\\ =3x+6 \)
I have not explained the initial factoring that I did in the first line. It is probably the hardest bit to understand.
Do you need me to explain more?? Which bits give you trouble.
Thank you!! It's wonderful to see you too! :)
That makes perfect sense, I forgot I had to factor first
Also next time I'll make sure to present it correctly
Thank you so much!