(1)/(5)(2y+3y)-8=-y+4

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Solution for (1)/(5)(2y+3y)-8=-y+4 equation:



(1)/(5)(2y+3y)-8=-y+4
We move all terms to the left:
(1)/(5)(2y+3y)-8-(-y+4)=0
Domain of the equation: 5(2y+3y)!=0
y∈R
We add all the numbers together, and all the variables
1/5(+5y)-(-1y+4)-8=0
We get rid of parentheses
1/5(+5y)+1y-4-8=0
We multiply all the terms by the denominator
1y*5(+5y)-4*5(+5y)-8*5(+5y)+1=0
Wy multiply elements
5y^2(+-20y(+-40y(++1=0
We use the square of the difference formula
5y^2(-20y(-40y(+1=0

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