68=1/5(20x+5-)+2

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Solution for 68=1/5(20x+5-)+2 equation: 



68=1/5(20x+5-)+2

We move all terms to the left:

68-(1/5(20x+5-)+2)=0



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

x∈R



We add all the numbers together, and all the variables

-(1/5(+20x)+2)+68=0

We multiply all the terms by the denominator


-(1+68*5(+20x)+2)=0



We calculate terms in parentheses: -(1+68*5(+20x)+2), so:

1+68*5(+20x)+2

determiningTheFunctionDomain
68*5(+20x)+1+2

We add all the numbers together, and all the variables

68*5(+20x)+3

Wy multiply elements

340x(++3

We use the square of the difference formula

340x(+3

Back to the equation:

-(340x(+3)



We add all the numbers together, and all the variables

-(340x3=0

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