1/2+(3z)/6=z/4-(2z+2)/3

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Solution for 1/2+(3z)/6=z/4-(2z+2)/3 equation: 


z in (-oo:+oo)

(3*z)/6+1/2 = z/4-((2*z+2)/3) // - z/4-((2*z+2)/3)

(2*z+2)/3+(3*z)/6-(z/4)+1/2 = 0

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

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

2*4*(2*z+2)+3*4*z+2*3*(-z)+1*3*4 = 0

28*z-6*z+12+16 = 0

22*z+12+16 = 0

22*z+28 = 0

(22*z+28)/(2*3*4) = 0

(22*z+28)/(2*3*4) = 0 // * 2*3*4

22*z+28 = 0

22*z+28 = 0 // - 28

22*z = -28 // : 22

z = -28/22

z = -14/11

z = -14/11

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