1(t-6)+7t=4(2t+3)-8

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Solution for 1(t-6)+7t=4(2t+3)-8 equation:



1(t-6)+7t=4(2t+3)-8
We move all terms to the left:
1(t-6)+7t-(4(2t+3)-8)=0
We add all the numbers together, and all the variables
7t+1(t-6)-(4(2t+3)-8)=0
We calculate terms in parentheses: -(4(2t+3)-8), so:
4(2t+3)-8
We multiply parentheses
8t+12-8
We add all the numbers together, and all the variables
8t+4
Back to the equation:
-(8t+4)
We get rid of parentheses
7t+1(t-6)-8t-4=0
We add all the numbers together, and all the variables
-1t+1(t-6)-4=0
We move all terms containing t to the left, all other terms to the right
-1t+1(t-6)=4

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