Question 82635
{{{3x + 4y <= 12}}}
{{{x + 3y <= 6}}}
{{{x >= 0}}}
{{{y >= 0}}}
Look at the 3rd and 4th first. {{{x >= 0}}} simply says that x is
never negative. {{{y >= 0}}} says y is never negative. The area
on the graph where both are true is the upper right quadrant, so
all solutions must be there as a restriction.
Solve the 1st and 2nd for y.
{{{3x + 4y <= 12}}}
{{{4y <= 12 - 3x}}}
{{{y <= 3 - (3/4)x}}}
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{{{x + 3y <= 6}}}
{{{3y <= 6 - x}}}
{{{y <= 2 - x/3}}}
now plot these
{{{ graph( 300, 300, -2, 8, -2, 8, 3 - (3/4)x, 2 - x/3) }}}
The solution is the area below the lines between (0,2) and (4,0)
Note that both these points are solutions and neither one
violates  {{{x >= 0}}} or {{{y >= 0}}}.