17b-3-12b=1/5*(25b-10)

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Solution for 17b-3-12b=1/5*(25b-10) equation: 


b in (-oo:+oo)

17*b-(12*b)-3 = (1/5)*(25*b-10) // - (1/5)*(25*b-10)

17*b-((1/5)*(25*b-10))-(12*b)-3 = 0

(-1/5)*(25*b-10)+17*b-12*b-3 = 0

17*b-1/5*(25*b-10)-12*b-3 = 0

17*b-1/5*(25*b-10)-12*b-3 = 0

12*b-12*b-3+2 = 0

2-3 = 0

-1 = 0

-1 = 0

b belongs to the empty set

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