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 Depdendent Variable

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 Dependent Variable

 Number of inequalities to solve: 23456789
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# Graphing Linear Inequalities

Like linear equations, linear inequalities start by graphing as a line. Unlike linear equalities this
line is a border between two regions that may contain more values that make the original
statement equation true. To further complicate matters the border itself may or may not be true.
The following examples will demonstrate these ideas.

Example 1: Graph x + y > 3 .

Solution

Step 1: Find the border.

The dividing line is found by setting the equation equal to zero and then plotting
several points to establish it.

 X intercept Y intercept Check point   Since the problem states that the equation is greater than but not equal to three,
this results in the line being “broken” and indicates that the line itself is not part
of the true solutions.

Step 2: Test points on both sides of the barrier.

Select a point that is not on the line and substitute its values into the equation. If
the answer is TRUE all the points in that region are true. If the answer is FALSE
all of the points in that region are false. Even if you find a true point to start
with, test a point from the opposite region to confirm your findings. For this
problem the origin (0, 0) is the easiest value to test. The point (1, 4) will be used
to confirm.

 x + y > 3 Let x & y = 0 0 + 0 > 3 0 > 3 (FALSE) x + y > 3 Let x = 1 & Let y = 4 1 + 4 > 3 5 > 3 (TRUE)

Now we graph the line for x + y = 3 and our two test points. Remember to use a
dashed line since the original equation was x + y > 3. Step 3: Shade the true areas

Since (0, 0) yielded a false answer and (1, 4) yielded a true answer, the region
that contained (1, 4) is shaded to indicate that any point in the region will give Example 2: Graph x + 2y ≤ 3.

Solution

Step 1: Find the barrier.

This problem is solved in the same manner as Ex. 1, but notice the equation is
looking for values less than or equal to 3. This allows the border to be included
as part of the solution, so in this case the border line will be solid, not broken.
The line will be solved for by using the x and y intercepts and x = -3, (-3, 3) as a
check point.

 X int. Y int. Check Pt.   Step 2: Test points on both sides of the barrier.

In this problem (0, 0) and (1, 4) will be tested. Step 3: Shade the true areas

Since (0, 0) is true
and (1, 4) is false,
the region
containing (0, 0) Example 3: Graph y < 4 .

Solution

Step 1: Find the barrier.

Recall that this equation represents a horizontal line. This means that the region
to be shaded will be above or below the line. Since the equation indicates that
the values are less than 4 and not equal to 4 the line is broken and not solid.

 y = 4 Let x = 0 (0, 4) y = 4 Let x = 1 Let x = 1 y = 4 Let x = - 1 (-1, 4)

Step 2: Test points on both sides of the barrier.

For this problem (0, 0) and (0, 5) will be tested.

 y < 4 Let y = 0 0 < 4 (TRUE) y < 4 Let y = 5 5 < 4 (FALSE) Step 3: Shade the true areas.

Since (0, 0) lies
in the true area and
(0, 5) does not, the
area with (0, 0) is Example 4: Graph x ≥ 2

Solution

Step 1: Find the barrier.

Like the last example, this line is a special case, graphing as a vertical line. This
means that the true values will be either on the left or right side of the line.

 x = 2 Let y = 0 (2, 0) x = 2 Let y = - 1 (2, -1) x = 2 Let y = 1 (2, 1)

Step 2: Test points on both sides of the barrier.

In this example (0, 0) and (3, 0) will be the test points.

x ≥ 2
Let x = 0
0 ≥ 2
(False)

x ≥ 2
Let x = 3
3 ≥ 2
(TRUE) Step 3: Shade the true areas.

Since (0, 0) is
false and (3, 0) is
true, the area
containing 