Question

Solve for n: 3n+2/7 = 1 - 2n-3/4

Answer 1

yes

Answer 2

lol @not-a-cop

Answer 3

First you need to get all the terms with "n" in them on one side.

Answer 4

hope u can do now ....ahead .....

Answer 5

yea dog i agree

Answer 6

This question is also multiple choice, here are the possible answers, A)-1/26 B)41/26 C)10/13 D)7/13

Answer 7

i think ur options r wrong ... well wait @whpalmer4 @austinL

Answer 8

yea that's why this make no sense to me @@shkrina

Answer 9

So, what should I pick for the answer? Even though none of them make sense.

Answer 10

I'm not at all convinced the options are wrong, I think the problem has likely been miswritten here...

Answer 11

@truu can you draw the problem exactly as it appears on your screen, paper, book, stone tablet, whatever?

Answer 12

When I solve out the answer I get -1/140 which is not an answer that is offered. I am of the same belief as @whpalmer4

Answer 13

sure @whpalmer4

Answer 14

Yeah, this the problem, I bet:

\[\frac{3n+2}{7} = 1-\frac{2n-3}{4}\]

Answer 15

The solution to that equation does appear in the list...

Answer 16

Solve for n: \[\frac{ 3n+2 }{ 7 }=1-\frac{ 2n-3 }{ 4 }\]

Answer 17

That's better!

Answer 18

girl ur question was wrong .... ,any ways ....

Answer 19

Yeah, okay, in the future, please give us the correct problem at the outset :-)

If you can't use the equation editor for the problem statement, use parentheses.

\[(3n+2)/7 = 1 - (2n-3)/4\]

Answer 20

sorry about that @shkrina

Answer 21

In general, just copying down all the symbols won't get you the same meaning, because position matters as well!

Answer 22

and ok @whpalmer4

Answer 23

its kk hope u can do it ... by taking LCM and then cross multiply ....

Answer 24

You aren't the first person to make this mistake on OpenStudy, you won't be the last, but maybe it can be the last time you do it :-)

Answer 25

@shkrina even easier would be to multiply by all of the denominators...

Answer 26

yea I wont mess up other equations again lol @whpalmer4

Answer 27

1- is their

Answer 28

"is there" :-)

that's okay:

\[\frac{3n+2}{7} = 1 - \frac{2n-3}{4}\]Now multiply by 7*4 = 28
\[\frac{28*(3n+2)}{7} = 28*1 - \frac{28(2n-3)}{4}\]\[4(3n+2) = 28-7(2n-3)\]\[12n+8 = 28 - 14n + 21\]

Answer 29

poof! no more fractions :-)

Answer 30

well that is same as LCM ////////

Answer 31

well, with fewer steps :-)

Answer 32

for one, you don't actually have to find the LCM (though numbers may be larger if the denominators are not relatively prime).

Answer 33

@Truu are you able to finish solving that to find the answer?

Answer 34

I'm kinda confused actually

Answer 35

about how I got there, or how to solve that remaining equation?

Answer 36

how to solve the remaining, I understand how you got there

Answer 37

\[12n+8=28−14n+21\]Let's get all the \(n\) terms on one side. Add \(14n\) to both sides to get rid of the term on the right side:
\[12n + 8 + 14n = 28 -14n + 21 + 14n\]\[26n + 8 = 28 + 21\]Now let's move that 8 to the right side by subtracting 8 from both sides\[26n+8-8=28+21-8\]\[26n = 41\]Now divide both sides by 26
\[\frac{26n}{26} = \frac{41}{26}\]\[n = \frac{41}{26}\]

Answer 38

Hope that's clear enough, I have to go!

Answer 39

ok yea I knew how to do some of that. Thank you for the help.