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

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

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

We move all terms to the left:

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

We multiply parentheses

7t-8t-(3(t+2)+2(t-1))+21+16=0


We calculate terms in parentheses: -(3(t+2)+2(t-1)), so:

3(t+2)+2(t-1)

We multiply parentheses

3t+2t+6-2

We add all the numbers together, and all the variables

5t+4

Back to the equation:

-(5t+4)


We add all the numbers together, and all the variables

-1t-(5t+4)+37=0

We get rid of parentheses

-1t-5t-4+37=0

We add all the numbers together, and all the variables

-6t+33=0

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

-6t=-33

t=-33/-6

t=5+1/2