1l2r+2(3/4r-1)=1/4r+6

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Solution for 1l2r+2(3/4r-1)=1/4r+6 equation: 


x in (-oo:+oo)

2*((3/4)*r-1)+1*l^2*r = (1/4)*r+6 // - (1/4)*r+6

2*((3/4)*r-1)+1*l^2*r-((1/4)*r)-6 = 0

2*((3/4)*r-1)+l^2*r+(-1/4)*r-6 = 0

2*(3/4*r-1)+l^2*r-r/4-6 = 0

(2*4*(3/4*r-1))/4+(4*l^2*r)/4-r/4+(-6*4)/4 = 0

2*4*(3/4*r-1)+4*l^2*r-r-6*4 = 0

6*r+4*l^2*r-r-24-8 = 0

4*l^2*r+5*r-24-8 = 0

4*l^2*r+5*r-32 = 0

(4*l^2*r+5*r-32)/4 = 0

(4*l^2*r+5*r-32)/4 = 0 // * 4

4*l^2*r+5*r-32 = 0

x belongs to the empty set

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