1/5(4+r)=1/2(40+10r)

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Solution for 1/5(4+r)=1/2(40+10r) equation: 



1/5(4+r)=1/2(40+10r)

We move all terms to the left:

1/5(4+r)-(1/2(40+10r))=0



Domain of the equation: 5(4+r)!=0

r∈R





Domain of the equation: 2(40+10r))!=0

r∈R



We add all the numbers together, and all the variables

1/5(r+4)-(1/2(10r+40))=0

We calculate fractions


(2r1/(5(r+4)*2(10r+40)))+(-5rr/(5(r+4)*2(10r+40)))=0



We calculate terms in parentheses: +(2r1/(5(r+4)*2(10r+40))), so:

2r1/(5(r+4)*2(10r+40))

We multiply all the terms by the denominator


2r1

We add all the numbers together, and all the variables

2r

Back to the equation:

+(2r)





We calculate terms in parentheses: +(-5rr/(5(r+4)*2(10r+40))), so:

-5rr/(5(r+4)*2(10r+40))

We multiply all the terms by the denominator


-5rr

Back to the equation:

+(-5rr)



We get rid of parentheses

2r-5rr=0

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