3a+4b=(1-2c)/d

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Solution for 3a+4b=(1-2c)/d equation: 


x in (-oo:+oo)

3*a+4*b = (1-(2*c))/d // - (1-(2*c))/d

3*a-((1-(2*c))/d)+4*b = 0

3*a-((1-2*c)/d)+4*b = 0

(-1*(1-2*c))/d+3*a+4*b = 0

(-1*(1-2*c))/d+(3*a*d)/d+(4*b*d)/d = 0

3*a*d-1*(1-2*c)+4*b*d = 0

2*c+3*a*d+4*b*d-1 = 0

(2*c+3*a*d+4*b*d-1)/d = 0

(2*c+3*a*d+4*b*d-1)/d = 0 // * d

2*c+3*a*d+4*b*d-1 = 0

x belongs to the empty set

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