4(2r-4)/7=4R-5/5

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Solution for 4(2r-4)/7=4R-5/5 equation: 


x in (-oo:+oo)

(4*(2*r-4))/7 = 4*R-(5/5) // - 4*R-(5/5)

(4*(2*r-4))/7-(4*R)+5/5 = 0

(4*(2*r-4))/7-4*R+5/5 = 0

(4*(2*r-4))/7-4*R+1 = 0

(4*(2*r-4))/7+(-4*7*R)/7+(1*7)/7 = 0

4*(2*r-4)-4*7*R+1*7 = 0

8*r-28*R-16+7 = 0

8*r-28*R-9 = 0

(8*r-28*R-9)/7 = 0

(8*r-28*R-9)/7 = 0 // * 7

8*r-28*R-9 = 0

x belongs to the empty set

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