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x+1/3x=12
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x+1/3x=12 *I assume you mean (x+1)/3x = 12 x+1 = 36x 1 = 35x 1/35 = x
\[x+\frac{1}{3x}=12\] \[\frac{3x^2+1}{3x}=12\] \[3x^2+1=36x\] \[3x^2-16x=-1\] \[(1\sqrt{3}x^2)-(4\sqrt{x})^2=-1\] \[(\sqrt{3}x-4\sqrt{x})(\sqrt{3}x+4\sqrt{x})\]
X is equal to 3.5 i think
x+1/3x=12 |*x \[x^2+\frac{1}{3}=12x\] can multiply both sides with 3 and get \[3x^2-36x+1=0\] And solve it.
I suppose you could be looking at x + 1/3 x = 12 4/3 x = 12 x = 9
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no I meant x + one-thirdx=12. Can u please do this problem again?
TBates go to chat!!!! plz!!!
So it is x + (1/3) x = 12? If so then my last post has the worked out solution
arthurfan i did it see above
3/3x+1/3x=12 4/3x=12 x=12*3/4 x=9
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