3(2t-3)+13=4(t-2)-(1t+2)

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Solution for 3(2t-3)+13=4(t-2)-(1t+2) equation: 



3(2t-3)+13=4(t-2)-(1t+2)

We move all terms to the left:

3(2t-3)+13-(4(t-2)-(1t+2))=0

We add all the numbers together, and all the variables

3(2t-3)-(4(t-2)-(t+2))+13=0

We multiply parentheses

6t-(4(t-2)-(t+2))-9+13=0



We calculate terms in parentheses: -(4(t-2)-(t+2)), so:

4(t-2)-(t+2)

We multiply parentheses

4t-(t+2)-8

We get rid of parentheses

4t-t-2-8

We add all the numbers together, and all the variables

3t-10

Back to the equation:

-(3t-10)



We add all the numbers together, and all the variables

6t-(3t-10)+4=0

We get rid of parentheses

6t-3t+10+4=0

We add all the numbers together, and all the variables

3t+14=0

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

3t=-14

t=-14/3

t=-4+2/3

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