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Question 729547: Hi there! I'm wondering what the best method would be to solve a system of three linear inequalities with only two variables each?
Example:
3x+5y ‹(or equal to) 180
5x+2y ‹(or equal to) 90
8x+3y ‹(or equal to) 240
My first instinct was to solve using the elimination method, but I don't know how to apply it to 3 equations.
I'm on my tablet, so I apologize that I can't use the less-than-or-equal-to symbols.
Any help is immensely appreciated!
Answer by KMST(5328)   (Show Source):
You can put this solution on YOUR website! The way to show the solution to  is graphing on a 2-dimensional graph.
(It's lucky you have only 2 variables, because 3 dimensional graphs would be hard to do).
The solution to each inequality would include the line represented by the equation with the equal sign, plus all the space to one side of that line.
The solution to the system is the space where the solutions to the 3 inequalities overlap (if they overlap).
 passes through the points
(0,36), with  -->  -->  --> 
and
(60,0) with  -->  -->  --> 
The graph of the line is  and  graphs as (
An easy way to decide which side of the line is the solution to the inequality is to use a user-friendly "test point" such as (0,0) that is not on the line.
Is (0,0), with  part of the solution?
It makes  so it is part of the solution, along with all the other points on the same side of the line.
Not all lines/inequalities are that easy to solve, but
 and  can be solved in the same manner.
 passes through (18,0) and (0,45);
 passes through (30,0) and (0,80), and
point (0,0) is part of the solution to and too.
Very conveniently, (0,0) is part of the solution to all 3 inequalities, so it is part of the solution to the system.
The 3 lines seemed so very easy to draw, and it was easy to see that the solution is the space to the left and below all 3 of the lines.
Unfortunately, the drawing that was so easy does not properly show the solution.
It does not show where the green and red lines intersect.
The scale needs to be changed, and the points used to draw the line will not be very helpful with the new scale.
I had to get a better way to calculate points on the lines:
 <-->  ;
 <-->  and
 <-->  .
I also had to solve  to find that those two lines intersect at (210,-480), to figure out how far to extend the scale.
 The solution is the shaded area, including the blackened parts of the colored lines.
On those blackened parts of the colored lines, either
 or  or 
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