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5(t-9)=10-3(t+5)
We move all terms to the left:
5(t-9)-(10-3(t+5))=0
We multiply parentheses
5t-(10-3(t+5))-45=0
We calculate terms in parentheses: -(10-3(t+5)), so:We get rid of parentheses
10-3(t+5)
determiningTheFunctionDomain -3(t+5)+10
We multiply parentheses
-3t-15+10
We add all the numbers together, and all the variables
-3t-5
Back to the equation:
-(-3t-5)
5t+3t+5-45=0
We add all the numbers together, and all the variables
8t-40=0
We move all terms containing t to the left, all other terms to the right
8t=40
t=40/8
t=5
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