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

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



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

We move all terms to the left:

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



Domain of the equation: (t+3)!=0

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

t!=-3

t∈R





Domain of the equation: (t-3)!=0

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

t!=3

t∈R



We calculate fractions


(1*(t-3))/((t+3)*(t-3))+(2t+6)/((t+3)*(t-3))-4=0



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

1*(t-3))/((t+3)*(t-3)

We multiply all the terms by the denominator


1*(t-3))

Back to the equation:

+(1*(t-3)))





We calculate terms in parentheses: +(2t+6)/((t+3)*(t-3)), so:

2t+6)/((t+3)*(t-3)

We multiply all the terms by the denominator


2t*((t+3)*(t-3)+6)

Back to the equation:

+(2t*((t+3)*(t-3)+6))

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