How do you solve #\frac { y } { 6} - \frac { 2y } { 5} = 1+ \frac { y } { 3}#?

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Answer 1 · 2018-06-04T02:03:04

#y=-30/17#

First, when you solve an algebraic equation with a fraction in it, you must always find the HCF (highest common factor) between them.

Therefore, in this case, the HCF between #6# (or also #3*2#), #5#, and #3# is #30# (in fact #3*2*5# since they are all prime numbers).

After that, you divide #30# by every denominator of the equation
and the number you get you multiply it by every numerator.

So

#y/6 * 5/5 - (2y)/5 * 6/6 = 1 * 30/30 + y/3 * 10/10#

#(5y)/5 - (2y)/5 = 30/30 + (10y)/10#

#(5y-12y)/30=(30+10y)/30 #

Then you multiply both sides by #30#. In this way, you get rid of #30#, so now you have an equation without fractions.

#5y-12y=30+10y#

You move all the number with #y# on one side,

#5y-12y-10y=30#

You procede with the addictions,

#-17y=30#

And finally

#y=-30/17#

If there is anything you don't understand, please tell me.