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jxharmon

Member

Registered: 2007-01-08

Posts: 104

please help with graphing linear inequalities

solve the following system of linear inequalities by graphing
3x + 4y <=12
x + 3y <=6
x >=0
y >=0
how do I graph this could you show me please

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: please help with graphing linear inequalities

Remember to logically "AND" the trueness of the four graphs together.
For first 3-4-12 equation, it is a diagonal border sloping down to the right
because x and y are both the same sign, due to symmetry with two axes.
The y changes slower than the x because 4x and 3x have to equalize.
So (0,3) and (4,0) are two points that certainly make sense and I got
them by just plugging in 0 and getting the other variable, but notice they
are off by 4 on the x, and 3 on the y, as predicted above.
Now shade in the TRUE area below and left of this line, and including the line.
That was 1st equation.

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: please help with graphing linear inequalities

Now 1-3-6 equation:  Again borderline slopes down and rightward.
y changes downward 3 times slower than x goes right, so the slope is -1/3, not very steep.
(0,2) and (6,0) follows this description, start at (0,2) and go down 2 and right 6 to (6,0).
It is a straight line due to x and y being to the one power.
Again shade down and left of the borderline including line.

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: please help with graphing linear inequalities

Now the x and y greater and equal to zero equations make the graph just be in the 1st quadrant.
The first quadrant means make a huge plus sign at the origin, and everything in the upper-right
square that goes on forever is the first quadrant.
The 2nd quadrant is the upper-left square.
The 3rd qudrant is the lower-left square region.
4th is lower-right.
So you want to keep the shaded regions that are only to the right and above the origin including the origin and the positive axes because they said greater or EQUAL to zero.

Hence, the answer is a right triangular region with a short triangle cut into it at about y = 6/5, which
you can get by combining the two lines.  3 times one line minus the other line and divide by 5 to get y.
Or you can just graph it with a ruler and estimate that point where the triangle cuts into the hypotenuse of the right triangle.  (The hypotenuse is the longest side of a right triangle, usually the diagonal long side.)

Last edited by John E. Franklin (2007-02-14 10:31:21)

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igloo myrtilles fourmis

John E. Franklin

Member

Registered: 2005-08-29

Posts: 3,588

Re: please help with graphing linear inequalities

Here's a graph that takes only 250 bytes or so.

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igloo myrtilles fourmis