(1/5)(9t-5)=(t+5/10)

Solution for (1/5)(9t-5)=(t+5/10) equation:

(1/5)(9t-5)=(t+5/10)

We move all terms to the left:

(1/5)(9t-5)-((t+5/10))=0


Domain of the equation: 5)(9t-5)!=0

t∈R


We add all the numbers together, and all the variables

(+1/5)(9t-5)-((+t+5/10))=0

We multiply parentheses ..

(+9t^2+1/5*-5)-((+t+5/10))=0

We calculate fractions


9t^2/()+(-((t-125))/()=0


We calculate terms in parentheses: +(-((t-125))/(), so:

-((t-125))/(

We multiply all the terms by the denominator


-((t-125))


We calculate terms in parentheses: -((t-125)), so:

(t-125)

We get rid of parentheses

t-125

Back to the equation:

-(t-125)


We get rid of parentheses

-t+125

We add all the numbers together, and all the variables

-1t+125

Back to the equation:

+(-1t+125)


We get rid of parentheses

9t^2/()-1t+125=0

We multiply all the terms by the denominator


9t^2-1t*()+125*()=0

We add all the numbers together, and all the variables

9t^2-1t*()=0