SOLUTION: Solve the following system of linear inequalities by graphing. 3x + 4y < 12 x + 3y < 6 x > 0 y > 0 Hello, I have no idea where to begin. Thank you

Algebra ->  Graphs -> SOLUTION: Solve the following system of linear inequalities by graphing. 3x + 4y < 12 x + 3y < 6 x > 0 y > 0 Hello, I have no idea where to begin. Thank you      Log On


   

Question 119868: Solve the following system of linear inequalities by graphing.
3x + 4y < 12
x + 3y < 6
x > 0
y > 0
Hello,
I have no idea where to begin.
Thank you
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
3x+%2B+4y+%3C+12
x+%2B+3y+%3C+6
x+%3E+0
y+%3E+0
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3x+%2B+4y+%3C+12
4y+%3C+-3x+%2B+12
y+%3C+-%283%2F4%29%2Ax+%2B+3
This means everything below (<)the line y+=+-%283%2F4%29%2Ax+%2B+3
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x+%2B+3y+%3C+6
3y+%3C+-x+%2B+6
y+%3C+-%281%2F3%29%2Ax+%2B+2
This means everything below the line y+=+-%281%2F3%29%2Ax+%2B+2
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You have to think about what the last two mean.
x+%3E+0 means x is never negative or zero. That would be
everthing to the right of (0,0), the center of the graph
------------------
y+%3E+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: