3(1+2s)=4(s-3)+19

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Solution for 3(1+2s)=4(s-3)+19 equation: 



3(1+2s)=4(s-3)+19

We move all terms to the left:

3(1+2s)-(4(s-3)+19)=0

We add all the numbers together, and all the variables

3(2s+1)-(4(s-3)+19)=0

We multiply parentheses

6s-(4(s-3)+19)+3=0



We calculate terms in parentheses: -(4(s-3)+19), so:

4(s-3)+19

We multiply parentheses

4s-12+19

We add all the numbers together, and all the variables

4s+7

Back to the equation:

-(4s+7)



We get rid of parentheses

6s-4s-7+3=0

We add all the numbers together, and all the variables

2s-4=0

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

2s=4

s=4/2

s=2

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