10(-7/2y)+10(1/5)=10(-3/2y)-10(6)

Solution for 10(-7/2y)+10(1/5)=10(-3/2y)-10(6) equation:

10(-7/2y)+10(1/5)=10(-3/2y)-10(6)

We move all terms to the left:

10(-7/2y)+10(1/5)-(10(-3/2y)-10(6))=0


Domain of the equation: 2y)!=0

y!=0/1

y!=0

y∈R



Domain of the equation: 2y)-106)!=0

y!=0/1

y!=0

y∈R


We add all the numbers together, and all the variables

10(-7/2y)-(10(-3/2y)-106)+10(+1/5)=0

We multiply parentheses

-70y-(10(-3/2y)-106)+1/5*10=0

We calculate fractions


-70y+()/2y-106)*5*10)+2y/2y-106)*5*10)=0

We calculate fractions


-70y+(()*2y)/4y^2+(-106)*5*10)+2y*2y)/4y^2=0

We multiply all the terms by the denominator


-70y*4y^2+(()*2y)+(-106)*5*10)+2y*2y)=0


We calculate terms in parentheses: +(()*2y), so:

()*2y


Wy multiply elements

-280y^3+(()*2y)+(-106)*5*10)+2y*2y)=0

We get rid of parentheses

-280y^3+(()*2y)-106)*5*10)+2y*2y=0


We calculate terms in parentheses: +(()*2y), so:

()*2y


We do not support eypression: y^3