5-2(z-4)/3=1/2(2z+3)

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Solution for 5-2(z-4)/3=1/2(2z+3) equation: 



5-2(z-4)/3=1/2(2z+3)

We move all terms to the left:

5-2(z-4)/3-(1/2(2z+3))=0



Domain of the equation: 2(2z+3))!=0

z∈R



We calculate fractions


(-2(z-4)*2(2z+3)))/(6z2+()/(6z2+5=0

We calculate fractions


((-2(z-4)*2(2z+3)))*(6z2+5)/((6z2*(6z2+5)+(()*6z2/((6z2*(6z2+5)=0



We calculate terms in parentheses: +((-2(z-4)*2(2z+3)))*(6z2+5)/((6z2*(6z2+5)+(()*6z2/((6z2*(6z2+5), so:

(-2(z-4)*2(2z+3)))*(6z2+5)/((6z2*(6z2+5)+(()*6z2/((6z2*(6z2+5

We can not solve this equation

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