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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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


We calculate terms in parentheses: -((4t+3)-2(t-4)), so:

(4t+3)-2(t-4)

We multiply parentheses

(4t+3)-2t+8

We get rid of parentheses

4t-2t+3+8

We add all the numbers together, and all the variables

2t+11

Back to the equation:

-(2t+11)


We add all the numbers together, and all the variables

-3t-(2t+11)-6=0

We get rid of parentheses

-3t-2t-11-6=0

We add all the numbers together, and all the variables

-5t-17=0

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

-5t=17

t=17/-5

t=-3+2/5