Question 362462
{{{(6x^2-5x-1)/(12x^2-11x+2)<0}}} 
{{{((x-1)(6x+1))/((3x-2)(4x-1))<0}}}
1, -1/6, 2/3, 1/4
Break up the number line into 5 regions using the critical points of the function.
Region 1:({{{-infinity}}},{{{-1/6}}})
Region 2:({{{-1/6}}},{{{1/4}}})
Region 3:({{{1/4}}},{{{2/3}}})
Region 4:({{{2/3}}},{{{1}}})
Region 5:({{{1}}},{{{infinity}}})
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For each region, choose a point in the region (not an endpoint).
Test the inequality.
If the inequality is satisfied, the region is part of the solution region.
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Region 1:{{{x=-1}}}
{{{((x-1)(6x+1))/((3x-2)(4x-1))<0}}}
{{{((-1-1)(-6+1))/((-3-2)(-4-1))<0}}}
{{{((-2)(-5))/((-5)(-5))<0 }}}
{{{2/5>0 }}}
False, Region 1 is not part of the solution region.
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Region 2:{{{x=0}}}
{{{((x-1)(6x+1))/((3x-2)(4x-1))<0}}}
{{{((-1)(1))/((-2)(-1))<0}}}
{{{-(1/2)< 0 }}}
True, Region 2 is part of the solution region.
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Region 3:{{{x=1/2}}}
{{{((x-1)(6x+1))/((3x-2)(4x-1))<0}}}
{{{((1/2-1)(3+1))/((3/2-2)(2-1))<0}}}
{{{((-1/2)(4))/((-1/2)(1))<0 }}}
{{{4<0 }}}
False, Region 3 is not part of the solution region.
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Region 4:{{{x=3/4}}}
{{{((x-1)(6x+1))/((3x-2)(4x-1))<0}}}
{{{((3/4-1)(9/2+1))/((9/4-2)(3-1))<0}}}
{{{((-1/4)(11/2))/((1/4)(2))<0 }}}
{{{-11/4<0 }}}
True, Region 4 is part of the solution region.
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Region 5:{{{x=2}}}
{{{((x-1)(6x+1))/((3x-2)(4x-1))<0}}}
{{{((2-1)(12+1))/((6-2)(8-1))<0}}}
{{{((1)(13))/((4)(7))<0 }}}
{{{13/28<0 }}}
False, Region 5 is not part of the solution region.
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Solution Region: ({{{-1/6}}},{{{1/4}}}) U ({{{2/3}}},{{{1}}})
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Graphical verification: Look for regions where the function is below the x-axis ({{{y<0}}})
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{{{drawing(300,300,-2,2,-2,5,grid(1),blue(line(-1/6,500,-1/6,-500)),blue(line(1,500,1,-500)),blue(line(2/3,500,2/3,-500)),blue(line(1/4,500,1/4,-500)),graph(300,300,-2,2,-2,5, ((x-1)(6x+1))/((3x-2)(4x-1)) ))}}}