Question 134247
{{{1/6-(1/3)t>0}}} Start with the given inequality




{{{(6)(1/cross(6)-(1/cross(3))t)=(6)(0)}}} Multiply both sides by the LCM of 6. This will eliminate the fractions  (note: if you need help with finding the LCM, check out this <a href=http://www.algebra.com/algebra/homework/divisibility/least-common-multiple.solver>solver</a>)




{{{1-2t=0}}} Distribute and multiply the LCM to each side




{{{-2t>0-1}}}Subtract 1 from both sides



{{{-2t>-1}}} Combine like terms on the right side



{{{t<(-1)/(-2)}}} Divide both sides by -2 to isolate t  (note: Remember, dividing both sides by a negative number flips the inequality sign) 




{{{t<1/2}}} Reduce


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Answer:

So our answer is {{{t<1/2}}}  (which is approximately {{{t<0.5}}} in decimal form)





So the answer in interval notation is *[Tex \Large \left(-\infty,\frac{1}{2}\right)]



And here's the graph of {{{t<1/2}}}



{{{drawing(500,50,-9,11,-10,10,
number_line(500,-9,11),
circle(1/2,-5.8,0.35),
circle(1/2,-5.8,0.4),
circle(1/2,-5.8,0.45),
blue(line(1/2,-5,1/2-10,-5)),
blue(line(1/2,-6,1/2-10,-6)),
blue(line(1/2,-7,1/2-10,-7)),
blue(arrow(1/2,-5,1/2-10.2,-5)),
blue(arrow(1/2,-5.5,1/2-10.2,-5.5)),
blue(arrow(1/2,-6,1/2-10.2,-6))
)}}}  Graph of {{{t<1/2}}} with the shaded region in blue


note: at the point {{{t=1/2}}}, there is an <font size=4><b>open</b></font> circle. This means the point {{{t=1/2}}} is excluded from the solution set.