(t+1)/2=(t-4)/5

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Solution for (t+1)/2=(t-4)/5 equation: 



(t+1)/2=(t-4)/5

We move all terms to the left:

(t+1)/2-((t-4)/5)=0

We calculate fractions


t/()+(-((t-4)*2)/()=0



We calculate terms in parentheses: +(-((t-4)*2)/(), so:

-((t-4)*2)/(

We multiply all the terms by the denominator


-((t-4)*2)



We calculate terms in parentheses: -((t-4)*2), so:

(t-4)*2

We multiply parentheses

2t-8

Back to the equation:

-(2t-8)



We get rid of parentheses

-2t+8

Back to the equation:

+(-2t+8)



We get rid of parentheses

t/()-2t+8=0

We multiply all the terms by the denominator


t-2t*()+8*()=0

We add all the numbers together, and all the variables

t-2t*()=0

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