5(6y+3)-19=3(4y-4)+22

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Solution for 5(6y+3)-19=3(4y-4)+22 equation: 



5(6y+3)-19=3(4y-4)+22

We move all terms to the left:

5(6y+3)-19-(3(4y-4)+22)=0

We multiply parentheses

30y-(3(4y-4)+22)+15-19=0



We calculate terms in parentheses: -(3(4y-4)+22), so:

3(4y-4)+22

We multiply parentheses

12y-12+22

We add all the numbers together, and all the variables

12y+10

Back to the equation:

-(12y+10)



We add all the numbers together, and all the variables

30y-(12y+10)-4=0

We get rid of parentheses

30y-12y-10-4=0

We add all the numbers together, and all the variables

18y-14=0

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

18y=14

y=14/18

y=7/9

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