8(1t+2)-3(1t-4)=6+(1t-7)+8

Solution for 8(1t+2)-3(1t-4)=6+(1t-7)+8 equation:

8(1t+2)-3(1t-4)=6+(1t-7)+8

We move all terms to the left:

8(1t+2)-3(1t-4)-(6+(1t-7)+8)=0

We add all the numbers together, and all the variables

8(t+2)-3(t-4)-(6+(t-7)+8)=0

We multiply parentheses

8t-3t-(6+(t-7)+8)+16+12=0


We calculate terms in parentheses: -(6+(t-7)+8), so:

6+(t-7)+8

determiningTheFunctionDomain
(t-7)+6+8

We add all the numbers together, and all the variables

(t-7)+14

We get rid of parentheses

t-7+14

We add all the numbers together, and all the variables

t+7

Back to the equation:

-(t+7)


We add all the numbers together, and all the variables

5t-(t+7)+28=0

We get rid of parentheses

5t-t-7+28=0

We add all the numbers together, and all the variables

4t+21=0

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

4t=-21

t=-21/4

t=-5+1/4