1/7(63x+42)=3x+20

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Solution for 1/7(63x+42)=3x+20 equation: 



1/7(63x+42)=3x+20

We move all terms to the left:

1/7(63x+42)-(3x+20)=0



Domain of the equation: 7(63x+42)!=0

x∈R



We get rid of parentheses

1/7(63x+42)-3x-20=0

We multiply all the terms by the denominator


-3x*7(63x+42)-20*7(63x+42)+1=0

Wy multiply elements

-21x^2(6-140x(6+1=0

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