40-3(x-5)=2/7(7x-35)

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Solution for 40-3(x-5)=2/7(7x-35) equation: 



40-3(x-5)=2/7(7x-35)

We move all terms to the left:

40-3(x-5)-(2/7(7x-35))=0



Domain of the equation: 7(7x-35))!=0

x∈R



We multiply parentheses

-3x-(2/7(7x-35))+15+40=0

We multiply all the terms by the denominator


-3x*7(7x-35))-(2+15*7(7x-35))+40*7(7x-35))=0

We add all the numbers together, and all the variables

-3x*7(7x-35))-(2+15*7(7x-35))+40*7(7x=0

Wy multiply elements

-21x^2(7+735x^2(7=0

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