Question 119868
{{{3x + 4y < 12}}}
{{{x + 3y < 6}}}
{{{x > 0}}}
{{{y > 0}}}
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{{{3x + 4y < 12}}}
{{{4y < -3x + 12}}}
{{{y < -(3/4)*x + 3}}}
This means everything below (<)the line {{{y = -(3/4)*x + 3}}}
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{{{x + 3y < 6}}}
{{{3y < -x + 6}}}
{{{y < -(1/3)*x + 2}}}
This means everything below the line {{{y = -(1/3)*x + 2}}}
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You have to think about what the last two mean.
{{{x > 0}}} means {{{x}}} is never negative or zero. That would be
everthing to the right of (0,0), the center of the graph
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{{{y > 0}}} means everthing above (0,0)
When you put these two together, you get
the upper right quadrant of the graph
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The solution has to be above both lines and only in the 
upper right quadrant. Here's the graph:
{{{ graph( 600, 600, -10, 10, -10, 10, -(3/4)x + 3, -(1/3)x + 2) }}}