(1/4c)+(2/7)=(2/7)(c+1)-(1/28c)

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Solution for (1/4c)+(2/7)=(2/7)(c+1)-(1/28c) equation: 


c in (-oo:+oo)

(1/4)*c+2/7 = (2/7)*(c+1)-((1/28)*c) // - (2/7)*(c+1)-((1/28)*c)

(1/4)*c-((2/7)*(c+1))+(1/28)*c+2/7 = 0

(-2/7)*(c+1)+(1/4)*c+(1/28)*c+2/7 = 0

c/4-2/7*(c+1)+c/28+2/7 = 0

(-2/7*4*7*28*(c+1))/(4*7*28)+(7*28*c)/(4*7*28)+(4*7*c)/(4*7*28)+(2*4*28)/(4*7*28) = 0

7*28*c-2/7*4*7*28*(c+1)+4*7*c+2*4*28 = 0

28*c-28*c-224+224 = 0

224-224 = 0

0 = 0

0/(4*7*28) = 0

0/(4*7*28) = 0 // * 4*7*28

0 = 0

c belongs to the empty set

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