5=2/7(10r+35)

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Solution for 5=2/7(10r+35) equation: 



5=2/7(10r+35)

We move all terms to the left:

5-(2/7(10r+35))=0



Domain of the equation: 7(10r+35))!=0

r∈R



We multiply all the terms by the denominator


-(2+5*7(10r+35))=0



We calculate terms in parentheses: -(2+5*7(10r+35)), so:

2+5*7(10r+35)

determiningTheFunctionDomain
5*7(10r+35)+2

Wy multiply elements

35r(1+2

Back to the equation:

-(35r(1+2)



We add all the numbers together, and all the variables

-(35r3=0

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