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

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

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

We move all terms to the left:

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

We multiply parentheses

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

We get rid of parentheses

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


We calculate terms in parentheses: -(6(t-3)-1), so:

6(t-3)-1

We multiply parentheses

6t-18-1

We add all the numbers together, and all the variables

6t-19

Back to the equation:

-(6t-19)


We add all the numbers together, and all the variables

5t-(6t-19)+6=0

We get rid of parentheses

5t-6t+19+6=0

We add all the numbers together, and all the variables

-1t+25=0

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

-t=-25

t=-25/-1

t=+25