solve for x: 3x-1/5 = 2x-2/3

[sportsaholic2397]

help me solve this and tell me if i am wrong

sorry i cant figure out the way to type in fractions...

3x-1/5 = 2x-2/3

start by multiplying the denominators by both top and bottom..

3x-1/5 (3/3) = 2x-2/3 (5/5)

9x-3/15 + 10x-10/15

combine

9x-3+10x-10/15

from here i 1, dont know if it is right and 2, not sure what to do from this point...

 

[Denis]

Re: solve for x

No. And messy :shock:
OK: what do you get if you multiply each term by 15?

 

[sportsaholic2397]

Re: solve for x

each term being, 9x-3 and 10x-10 ?

it would come out to be 135x-45 and 150x-150.

it has been a while and i am trying to get the kinks out for this online class...

 

[Deleted member 4993]

Re: solve for x

help me solve this and tell me if i am wrong

sorry i cant figure out the way to type in fractions...

3x-1/5 = 2x-2/3

- should have been written as (3x - 1)/5 = (2x-2)/3

\(\displaystyle \frac{3x-1}{5} \, = \, \frac{2x-2}{3}\)

multiply both sides by 15 (lowest-common-multiple of the denominators) and you get

\(\displaystyle 3\cdot (3x-1) \, = \, 5\cdot (2x-2)\)

\(\displaystyle 9x-3 \, = \, 10x-10\)

Isolate 'x' - that is bring all the terms with 'x' on one side of the "=" sign and rest on the other side

\(\displaystyle -3 + 10\, = \, 10x-9x\)

Now finish it


start by multiplying the denominators by both top and bottom..

3x-1/5 (3/3) = 2x-2/3 (5/5)

9x-3/15 + 10x-10/15

combine

9x-3+10x-10/15

from here i 1, dont know if it is right and 2, not sure what to do from this point...

 

[Denis]

Re: solve for x

NO!! The original terms; you have:
(3x-1) / 5 = (2x-2) / 3 ...by the way, the brackets are IMPORTANT!

Multiply by 15; that'll give you:
3(3x - 1) = 5(2x - 2)

Ya betta get it now, suckah, or I'll punch your lights out :wink:

 

[Deleted member 4993]

Re: solve for x

NO!! ...
Ya betta get it now, suckah, or I'll punch your lights out :wink:

I thought you are about to get me - I ducked....