(1/9)(9t-5)=(t+5/18)

Solution for (1/9)(9t-5)=(t+5/18) equation:

(1/9)(9t-5)=(t+5/18)

We move all terms to the left:

(1/9)(9t-5)-((t+5/18))=0


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

t∈R


We add all the numbers together, and all the variables

(+1/9)(9t-5)-((+t+5/18))=0

We multiply parentheses ..

(+9t^2+1/9*-5)-((+t+5/18))=0

We calculate fractions


9t^2/()+(-((t-225))/()=0


We calculate terms in parentheses: +(-((t-225))/(), so:

-((t-225))/(

We multiply all the terms by the denominator


-((t-225))


We calculate terms in parentheses: -((t-225)), so:

(t-225)

We get rid of parentheses

t-225

Back to the equation:

-(t-225)


We get rid of parentheses

-t+225

We add all the numbers together, and all the variables

-1t+225

Back to the equation:

+(-1t+225)


We get rid of parentheses

9t^2/()-1t+225=0

We multiply all the terms by the denominator


9t^2-1t*()+225*()=0

We add all the numbers together, and all the variables

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