z=4/5(2c+a)

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Solution for z=4/5(2c+a) equation: 


x in (-oo:+oo)

z = (4/5)*(a+2*c) // - (4/5)*(a+2*c)

z-((4/5)*(a+2*c)) = 0

(-4/5)*(a+2*c)+z = 0

z-4/5*(a+2*c) = 0

z-4/5*(a+2*c) = 0

z-4/5*a-8/5*c = 0

z-4/5*a-8/5*c = 0

x belongs to the empty set

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