7-(4+3t)+2t=6(t-2)-5

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Solution for 7-(4+3t)+2t=6(t-2)-5 equation: 



7-(4+3t)+2t=6(t-2)-5

We move all terms to the left:

7-(4+3t)+2t-(6(t-2)-5)=0

We add all the numbers together, and all the variables

-(3t+4)+2t-(6(t-2)-5)+7=0

We add all the numbers together, and all the variables

2t-(3t+4)-(6(t-2)-5)+7=0

We get rid of parentheses

2t-3t-(6(t-2)-5)-4+7=0



We calculate terms in parentheses: -(6(t-2)-5), so:

6(t-2)-5

We multiply parentheses

6t-12-5

We add all the numbers together, and all the variables

6t-17

Back to the equation:

-(6t-17)



We add all the numbers together, and all the variables

-1t-(6t-17)+3=0

We get rid of parentheses

-1t-6t+17+3=0

We add all the numbers together, and all the variables

-7t+20=0

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

-7t=-20

t=-20/-7

t=2+6/7

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