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

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

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

We move all terms to the left:

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


Domain of the equation: 5(2+t)!=0

t∈R



Domain of the equation: 2(t-1))!=0

t∈R


We add all the numbers together, and all the variables

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

We calculate fractions


(2tt/(5(t+2)*2(t-1)))+(-5tt/(5(t+2)*2(t-1)))=0


We calculate terms in parentheses: +(2tt/(5(t+2)*2(t-1))), so:

2tt/(5(t+2)*2(t-1))

We multiply all the terms by the denominator


2tt

Back to the equation:

+(2tt)



We calculate terms in parentheses: +(-5tt/(5(t+2)*2(t-1))), so:

-5tt/(5(t+2)*2(t-1))

We multiply all the terms by the denominator


-5tt

Back to the equation:

+(-5tt)


We get rid of parentheses

2tt-5tt=0