A community for students.
Here's the question you clicked on:
 0 viewing

This Question is Closed

nbouscal
 3 years ago
Best ResponseYou've already chosen the best response.1I would start by multiplying both sides of the equation by 3x+9 and see what happens :)

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks, I'll see if that gets me the answer.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1I have a suggestion here. Shall I spit it out?

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0Nah, don't work, it's a fillin answer. @ParthKohli Sure.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Multiply both the numerator and denominator of \(\Large {2 \over x + 3}\).That'll get you two like terms :)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Multiply both numerator and denominator by 3*

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0okay, bottom is 9, top is 6. Is there another step?

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0I do see the nine and nine, like terms.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1\(\color{Black}{\Rightarrow \Large{1 \over 3x + 9} {3(2) \over 3(x + 3)} }\)

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0I see, but still confusing.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1I don't find anything confusing here. \(\color{Black}{\Rightarrow \Large {x \over y} = {ax \over ay}}\) This is what I am doing here.

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0I know, but this is one of the first units, I'm really, for some reason, having trouble with. Okay, you say, 'Multiply both numerator and denominator by 3*', Okay then what do I do?

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1You'll get two like terms :)

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1First simplify the thing I gave you

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0\[1/3x + 9  6 / x9\] ^^Don't know how to really format it, but the 6 and 9 is there.

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0Or am I missing something

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1\(\color{Black}{\Rightarrow 3(x + 3) = 3x + 9 }\)

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0okay look, we both are going in circles, but probably my fault, but I see the like terms, but what do I do now? :L lol...

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1just operate on the numerators.

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0Nope, couldn't get anything. Okay, I don't want to waste anymore of your precious time, so I'ma ask for a straight up answer? This pellet is just hard for me, point blank.

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0or tell me next steps, I don't mind learning it.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1Wolfram if you want the straight answer.

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0lol i just want to learn the steps, but I don't want to waste ur time tutoring me.

ParthKohli
 3 years ago
Best ResponseYou've already chosen the best response.1\(\color{Black}{\Rightarrow \Large {1 \over 3x+ 9} {6 \over 3x + 9} = 2 }\) \(\color{Black}{\Rightarrow \Large {5 \over 3x + 9} = 2 }\) Multiply both sides by 3x + 9 \(\color{Black}{\Rightarrow5 = 2(3x+ 9) }\) \(\color{Black}{\Rightarrow 5 = 6x + 18 }\) \(\color{Black}{\Rightarrow 23 = 6x }\) \(\color{Black}{\Rightarrow x = {23 \over 6} }\)

FeoNeo33
 3 years ago
Best ResponseYou've already chosen the best response.0k i was good up till the part with the Multiply both sides by 3x + 9... but Ima let you go, I promise, Im not here just for the answer. Ima learn this stuff, I don't want to be a leecher my whole life. Thanks, bye.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.