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4-2(2z-4)/3=1/2(2z+5)
We move all terms to the left:
4-2(2z-4)/3-(1/2(2z+5))=0
Domain of the equation: 2(2z+5))!=0We calculate fractions
z∈R
(-2(2z-4)*2(2z+5)))/(6z2+()/(6z2+4=0
We calculate fractions
((-2(2z-4)*2(2z+5)))*(6z2+4)/((6z2*(6z2+4)+(()*6z2/((6z2*(6z2+4)=0
We calculate terms in parentheses: +((-2(2z-4)*2(2z+5)))*(6z2+4)/((6z2*(6z2+4)+(()*6z2/((6z2*(6z2+4), so:
(-2(2z-4)*2(2z+5)))*(6z2+4)/((6z2*(6z2+4)+(()*6z2/((6z2*(6z2+4
We can not solve this equation
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