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5(4+3/2y)+4y=10
We move all terms to the left:
5(4+3/2y)+4y-(10)=0
Domain of the equation: 2y)!=0We add all the numbers together, and all the variables
y!=0/1
y!=0
y∈R
5(3/2y+4)+4y-10=0
We add all the numbers together, and all the variables
4y+5(3/2y+4)-10=0
We multiply parentheses
4y+15y+20-10=0
We add all the numbers together, and all the variables
19y+10=0
We move all terms containing y to the left, all other terms to the right
19y=-10
y=-10/19
y=-10/19
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