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1/7(3(x-1)+35)=x
We move all terms to the left:
1/7(3(x-1)+35)-(x)=0
Domain of the equation: 7(3(x-1)+35)!=0We add all the numbers together, and all the variables
x∈R
-1x+1/7(3(x-1)+35)=0
We multiply all the terms by the denominator
-1x*7(3(x-1)+35)+1=0
Wy multiply elements
-7x^2(3+1=0
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