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5x+1/4(12x+28)=4(2x+1)+3
We move all terms to the left:
5x+1/4(12x+28)-(4(2x+1)+3)=0
Domain of the equation: 4(12x+28)!=0We multiply all the terms by the denominator
x∈R
5x*4(12x+28)-((4(2x+1)+3))*4(12x+28)+1=0
We calculate terms in parentheses: -((4(2x+1)+3))*4(12x+28), so:Wy multiply elements
(4(2x+1)+3))*4(12x+28
20x^2(1-((4(2x+1)+3))*4(12x+28)+1=0
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