6y+4-2(1-y)=10(y+2)-2(y+3)

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Solution for 6y+4-2(1-y)=10(y+2)-2(y+3) equation: 



6y+4-2(1-y)=10(y+2)-2(y+3)

We move all terms to the left:

6y+4-2(1-y)-(10(y+2)-2(y+3))=0

We add all the numbers together, and all the variables

6y-2(-1y+1)-(10(y+2)-2(y+3))+4=0

We multiply parentheses

6y+2y-(10(y+2)-2(y+3))-2+4=0



We calculate terms in parentheses: -(10(y+2)-2(y+3)), so:

10(y+2)-2(y+3)

We multiply parentheses

10y-2y+20-6

We add all the numbers together, and all the variables

8y+14

Back to the equation:

-(8y+14)



We add all the numbers together, and all the variables

8y-(8y+14)+2=0

We get rid of parentheses

8y-8y-14+2=0

We add all the numbers together, and all the variables

-12!=0

There is no solution for this equation

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