15(t-5)+6t=7(3t+t)-11

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Solution for 15(t-5)+6t=7(3t+t)-11 equation:



15(t-5)+6t=7(3t+t)-11
We move all terms to the left:
15(t-5)+6t-(7(3t+t)-11)=0
We add all the numbers together, and all the variables
15(t-5)+6t-(7(+4t)-11)=0
We add all the numbers together, and all the variables
6t+15(t-5)-(7(+4t)-11)=0
We multiply parentheses
6t+15t-(7(+4t)-11)-75=0
We calculate terms in parentheses: -(7(+4t)-11), so:
7(+4t)-11
We multiply parentheses
28t-11
Back to the equation:
-(28t-11)
We add all the numbers together, and all the variables
21t-(28t-11)-75=0
We get rid of parentheses
21t-28t+11-75=0
We add all the numbers together, and all the variables
-7t-64=0
We move all terms containing t to the left, all other terms to the right
-7t=64
t=64/-7
t=-9+1/7

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