Question

I have a question.
When i get left with the equation (for example) 13n+8=-11n+11 , how do i know which side of the variable I subtract/add first? The 13n or -11n ?

Answer 1

it doesn't matter
first decide... do you want n by itself at left side or right side ???

Answer 2

but it is always good to keep the variable at left side :=))
o^_^o

Answer 3

Thanks. It's just sometimes i do it where i think it's right & end up with the wrong answer

Answer 4

hmm would you like to try for this equation ???
lets see if you get it right or not :=))

Answer 5

Yeah sure

Answer 6

I got n=3

Answer 7

did you subtract 13n or add 11n ??

Answer 8

Added 13n to 11n

Answer 9

When i did it your way, i got 3/-2n

Answer 10

can you post your entire work? pls

Answer 11

lets say you want `n` at left side what would be the first step ?

Answer 12

From the beginning of the actual problem? Or just from that piece I posted?

Answer 13

The actual problem is kinda long

Answer 14

how did you solve for n from there^^^?

Answer 15

Sorry i cant see what you pic you posted because I'm on my phone

Answer 16

I couldn't say the answer better myself ^_^

Answer 17

So from here: \(\large 13n + 8 = -11n + 11\)

Lets say we want to get 'n' alone on the right side of the equation

In order to get rid of all the *n's* on the left side...we need to get rid of that 13n.how would we do that?

I noticed before you said you ADDED 13n to both sides...if we do that...we just get:

\[\large 26n + 8 = 2n + 11\]
That didn't get rid of anything we needed....we actually needed to SUBTRACT 13n from both sides...because notice when you do that...on the left side you have 13n - 13n which then cancels it out

So subtracting 13n from both sides...we get:
\[\large 8 = -24n + 11\]

Then when we isolate the 'n' ***by subtracting 11 from both sides and then dividing by -24***...we arrive at the answer \(\large n = \frac{1}{8}\)

Try it out by getting 'n' alone on the left side this time..see if you arrive at the same result :)

Answer 18

Here is one piece of advice. I would do it, if possible, so that you get the positive on one side.

Example:

\(2x+3=3x+1\)

I always start with the variables for some reason

I can either subtract \(2x\) from both sides or \(3x\) from both sides. The first way I end up with a positive amount of \(x'\)s and the second way we have a negative amount.

People tend to make more mistakes when they have a negative number in front of a variable, so I think this can save some people some agony. :)

Answer 19

that's true^^

Answer 20

Another note of importance. You can always switch things to either side of an equation. Most people like to stay on the left side, but there is no reason to do so.

If \(2x+3=3x+1\) then you can always do this \(3x+1=2x+3\).

Equals sign works both ways!

Answer 21

Fine tuning: When you combine those x terms, try to do so in a way that produces a positive result. Less chance of error that way.

"13n+8=-11n+11": I would ADD 11n to both sides of the equation, thus producing +24n, and then I'd subtract 8 from both sides. Please do this. Your answer?

Answer 22

If you got -24n, that would not necessarily be wrong, but my strong opinion is that it's better to give the unknown quantity a positive sign.