c=(6/5)*(k-35)

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Solution for c=(6/5)*(k-35) equation: 


x in (-oo:+oo)

c = (6/5)*(k-35) // - (6/5)*(k-35)

c-((6/5)*(k-35)) = 0

(-6/5)*(k-35)+c = 0

c-6/5*(k-35) = 0

c-6/5*(k-35) = 0

c-6/5*k+42 = 0

c-6/5*k+42 = 0

x belongs to the empty set

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