3/4(8z+2)=1/6(36z+12(

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Solution for 3/4(8z+2)=1/6(36z+12( equation: 



3/4(8z+2)=1/6(36z+12(

We move all terms to the left:

3/4(8z+2)-(1/6(36z+12()=0



Domain of the equation: 4(8z+2)!=0

z∈R





Domain of the equation: 6(36z+12()!=0

z∈R



We calculate fractions


(18z3/(4(8z+2)*6(36z+12())+(-4z8/(4(8z+2)*6(36z+12())=0



We calculate terms in parentheses: +(18z3/(4(8z+2)*6(36z+12())+(-4z8/(4(8z+2)*6(36z+12()), so:

18z3/(4(8z+2)*6(36z+12())+(-4z8/(4(8z+2)*6(36z+12()

We can not solve this equation

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