4(t+3)-3(6-5t)=18(t-1)+2

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Solution for 4(t+3)-3(6-5t)=18(t-1)+2 equation: 



4(t+3)-3(6-5t)=18(t-1)+2

We move all terms to the left:

4(t+3)-3(6-5t)-(18(t-1)+2)=0

We add all the numbers together, and all the variables

4(t+3)-3(-5t+6)-(18(t-1)+2)=0

We multiply parentheses

4t+15t-(18(t-1)+2)+12-18=0



We calculate terms in parentheses: -(18(t-1)+2), so:

18(t-1)+2

We multiply parentheses

18t-18+2

We add all the numbers together, and all the variables

18t-16

Back to the equation:

-(18t-16)



We add all the numbers together, and all the variables

19t-(18t-16)-6=0

We get rid of parentheses

19t-18t+16-6=0

We add all the numbers together, and all the variables

t+10=0

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

t=-10

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