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(6/5)-(5x-3)=(1/4)x-12
We move all terms to the left:
(6/5)-(5x-3)-((1/4)x-12)=0
Domain of the equation: 4)x-12)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
-(5x-3)-((+1/4)x-12)+(+6/5)=0
We get rid of parentheses
-5x-((+1/4)x-12)+3+6/5=0
We calculate fractions
-5x+()/4x-12)*5)+24x/4x-12)*5)+3=0
We calculate fractions
-5x+(()*4x)/16x^2+(-12)*5)+24x*4x)/16x^2=0
We multiply all the terms by the denominator
-5x*16x^2+(()*4x)+(-12)*5)+24x*4x)=0
We calculate terms in parentheses: +(()*4x), so:Wy multiply elements
()*4x
-80x^3+(()*4x)+(-12)*5)+24x*4x)=0
We get rid of parentheses
-80x^3+(()*4x)-12)*5)+24x*4x=0
We calculate terms in parentheses: +(()*4x), so:We do not support expression: x^3
()*4x
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