(Y+4)+(2y)+(3y-1)=19/5

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Solution for (Y+4)+(2y)+(3y-1)=19/5 equation: 



(+4)+(2Y)+(3Y-1)=19/5

We move all terms to the left:

(+4)+(2Y)+(3Y-1)-(19/5)=0

We add all the numbers together, and all the variables

2Y+(3Y-1)+4-(+19/5)=0

We get rid of parentheses

2Y+3Y-1+4-19/5=0

We multiply all the terms by the denominator


2Y*5+3Y*5-19-1*5+4*5=0

We add all the numbers together, and all the variables

2Y*5+3Y*5-4=0

Wy multiply elements

10Y+15Y-4=0

We add all the numbers together, and all the variables

25Y-4=0

We move all terms containing Y to the left, all other terms to the right

25Y=4

Y=4/25

Y=4/25

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