1/2(2r+6)/5=-7x-4+7x

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Solution for 1/2(2r+6)/5=-7x-4+7x equation: 


x in (-oo:+oo)

((1/2)*(2*r+6))/5 = 7*x-7*x-4 // - 7*x-7*x-4

((1/2)*(2*r+6))/5-(-7*x)-(7*x)+4 = 0

((1/2)*(2*r+6))/5+7*x-7*x+4 = 0

(1/2*(2*r+6))/5+7*x-7*x+4 = 0

(1/2*(2*r+6))/5+(5*7*x)/5+(-7*5*x)/5+(4*5)/5 = 0

1/2*(2*r+6)+5*7*x-7*5*x+4*5 = 0

r+35*x-35*x+3+20 = 0

r+3+20 = 0

r+23 = 0

(r+23)/5 = 0

(r+23)/5 = 0 // * 5

r+23 = 0

x belongs to the empty set

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