How do you solve #1/3y + 2/5 - 2/3y= 4/8#?

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Answer 1 · 2014-11-05T17:18:32

#1/3y+2/5-2/3y=4/8#

by #4/8=1/2#,

#=> 1/3y+2/5-2/3y=1/2#

by subtracting #2/5#,

#=> 1/3y-2/3y=1/2-2/5#

by finding the common denominator (10),

#=> -1/3y=5/10-4/10={5-4}/10=1/10#

by multiply by #-3#,

#=>y=-3/10#.

I hope that this was helpful.