8(t-3)+4t=6(t+1)-10

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Solution for 8(t-3)+4t=6(t+1)-10 equation: 



8(t-3)+4t=6(t+1)-10

We move all terms to the left:

8(t-3)+4t-(6(t+1)-10)=0

We add all the numbers together, and all the variables

4t+8(t-3)-(6(t+1)-10)=0

We multiply parentheses

4t+8t-(6(t+1)-10)-24=0



We calculate terms in parentheses: -(6(t+1)-10), so:

6(t+1)-10

We multiply parentheses

6t+6-10

We add all the numbers together, and all the variables

6t-4

Back to the equation:

-(6t-4)



We add all the numbers together, and all the variables

12t-(6t-4)-24=0

We get rid of parentheses

12t-6t+4-24=0

We add all the numbers together, and all the variables

6t-20=0

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

6t=20

t=20/6

t=3+1/3

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