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5(1/5x+2)=4(x-2)
We move all terms to the left:
5(1/5x+2)-(4(x-2))=0
Domain of the equation: 5x+2)!=0We multiply parentheses
x∈R
5x-(4(x-2))+10=0
We calculate terms in parentheses: -(4(x-2)), so:We get rid of parentheses
4(x-2)
We multiply parentheses
4x-8
Back to the equation:
-(4x-8)
5x-4x+8+10=0
We add all the numbers together, and all the variables
x+18=0
We move all terms containing x to the left, all other terms to the right
x=-18
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