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David Cohen questions just for fun.
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y=sqrt(3)(1-3x)/3............1)
sqrt(3)y+3x -1=0..........2)
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sqrt(3)y+3x -1=0..........2), solve for y
sqrt(3)y=1-3x
y=(1-3x)/sqrt(3)............rationalize
y=sqrt(3)(1-3x)/3................same line as in 1)
=>
sqrt(3)(1-3x)/3=sqrt(3)(1-3x)/3
1-3x=1-3x
1-1=3x-3x
0=0
=>
The solutions to the system of equations are: y=sqrt(3)(1-3x)/3