63=1/5(20x+50)+2

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Solution for 63=1/5(20x+50)+2 equation: 



63=1/5(20x+50)+2

We move all terms to the left:

63-(1/5(20x+50)+2)=0



Domain of the equation: 5(20x+50)+2)!=0

x∈R



We multiply all the terms by the denominator


-(1+63*5(20x+50)+2)=0



We calculate terms in parentheses: -(1+63*5(20x+50)+2), so:

1+63*5(20x+50)+2

determiningTheFunctionDomain
63*5(20x+50)+1+2

We add all the numbers together, and all the variables

63*5(20x+50)+3

Wy multiply elements

315x(2+3

Back to the equation:

-(315x(2+3)



We add all the numbers together, and all the variables

-(315x5=0

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