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

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Answer 1 · 2018-06-24T10:03:05

#color(red)(y = 2#

#4y - ((3y + 4)/2) + 1/3 = 5 ((4-y)/3)#

#4y - (3y)/2 -2 + 1/3 = 20/3 - (5y)/3, " removing braces"#

#4y - (3y)/2 + (5y) / 3 = 20/3 + 2 - 1/3, "rearranging like terms together"#

#(24y - 9y + 10y) / 6 = (20 + 6 - 1) / 3, " taking L C M"#

#(25 y ) / cancel(6)^color(red)(2) = 25 / cancel3, " simplifying"#

#25y = 25 * 2 = 50, " cross multiplying"#

#color(red)(y = 50 / 25 = 2#