A.REI.12: Graph Systems of Inequalities
A.REI.12: Graph Systems of Inequalities
Represent and solve equations and inequalities graphically.
12. Graph the solutions to a linear inequality in two variables as a half-plane (excluding the boundary in the case of a strict
inequality), and graph the solution set to a system of linear inequalities in two variables as the intersection of the
corresponding half-planes.
Overview of Lesson
- activate prior knowledge
- present vocabulary and/or big ideas associated with the lesson
- connect assessment practices with curriculum
- model an assessment problem and solution strategy
- facilitate guided practice of student activity
Optional: Provide or allow students to create additional problem sets
- facilitate a summary and share out of student work
Optional HW - Write the math assignment.
Big Ideas
A linear inequality describes a region of the coordinate plane that has a boundary line. Every point in
the region is a solution of the inequality.
Two or more linear inequalities together form a system of linear inequalities. Note that there are two
or more boundary lines in a system of linear inequalities.
A solution of a system of linear inequalities makes each inequality in the system true. The graph of a
system shows all of its solutions.
Graphing a Linear Inequality
Step One. Change the inequality sign to an equal sign and graph the boundary line in the same manner
that you would graph a linear equation.
When the inequality sign contains an equality bar beneath it, use a solid line for the boundary.
When the inequality sign does not contain an equality bar beneath it, use a dashed or dotted line
for the boundary
Step Two. Restore the inequality sign and test a point to see which side of the boundary line the
solution is on. The point (0,0) is a good point to test since it simplifies any multiplication. However, if
the boundary line passes through the point (0,0), another point not on the boundary line must be selected
for testing.
If the test point makes the inequality true, shade the side of the boundary line that includes the test
point.
If the test point makes the inequality not true, shade the side of the boundary line does not include
the test point.
Example Graph y  2 x  3
First, change the inequality sign an equal sign and graph the line: y  2 x  3 . This is the boundary line
of the solution. Since there is no equality line beneath the inequality symbol, use a dashed line for the
boundary.
Next, test a point to see which side of the boundary line the solution is on. Try (0,0), since it makes the
multiplication easy, but remember that any point will do.
y  2x  3
0  2(0)  3
03
True, so the solution of the inequality is the region that contains the point  0,0  .
Therefore, we shade the side of the boundary line that contains the point (0,0).
Note: The TI-83+ graphing calculator does not have the ability to distinguish between solid and dashed
lines on a graph of an inequality. The less than and greater than symbols are input using the far-left
column of symbols that can be accessed through the Y= feature.
Graphing a System of Linear Inequalities. Systems of linear inequalities are graphed in the same
manner as systems of equations are graphed. The solution of the system of inequalities is the region of
the coordinate plane that is shaded by both inequalities.
4 y  6x
Example: Graph the system:
3x  6 y  6
First, convert both inequalities to slope-intercept form and graph.
3x  6 y  6
4 y  6x
4 y 6x

4
4
3
y x
2
3
m= , b=0
2
D2
6 y  6  3x
6 y  3x  6
6 y 3x 6
 
6
6 6
1
y  x 1
2
1
m= , b=-1
2
Next, test a point in each inequality and shade appropriately.
3
Since point (0,0) is on the boundary line of y  x , select another point, such as (0,1).
2
3
x
2
Test (0,1)
3
1   0
2
1  0 This is true, so the point (0,1) is in the solution set of this inequality.
Therefore, we shade the side of the boundary line that includes point (0,1).
y
Since (0,0) is not on the boundary line of y 
1
x  1 , we can use (0,0) as our test point, as
2
follows:
1
y  x 1
2
Test (0,0)
1
 0 1
2
0  1 This is not true, so the point (0,0) is not in the solution set of this inequality.
We therefore must shade the side of the boundary line that does not incude the point (0,0).
0
Note that the system of inequalities divides the coordinate plane into four sections. The solution set for
the system of inequalities is the area where the two shaded regions overlap.
Remember The Big Rule for Solving Inequalities:
All the rules for solving equations apply to inequalties – plus one:
When an inequality is multiplied or divided by any negative number, the direction of the inequality
sign changes.
REGENTS PROBLEMS
1. What is one point that lies in the solution set of the system of inequalities graphed below?
a.
b.
c.
d.
2. Given:
Which graph shows the solution of the given set of inequalities?
a.
c.
b.
d.
3. Which ordered pair is not in the solution set of
a.
b.
4. Which graph represents the solution of
a.
b.
and
c.
d.
and
c.
d.
5. Which inequality is represented in the graph below?
a.
?
c.
?
b.
d.
6. The graph of an inequality is shown below.
a) Write the inequality represented by the graph.
b) On the same set of axes, graph the inequality
.
c) The two inequalities graphed on the set of axes form a system. Oscar thinks that the point
is in the
solution set for this system of inequalities. Determine and state whether you agree with Oscar. Explain your
reasoning.
7. On the set of axes below, graph the inequality
.
A.REI.12: Graph Systems of Inequalities
Answer Section
1. ANS: A
Strategy: Visually estimate whether a point falls in the solution area and eliminate wrong answers.
a.
b.
c.
d.
(7,0) clearly falls in the solutions area for both the solid line and the dotted line.
(3,0) appears to be in the solution area for the solid line, but not for the dotted line.
(0,7) is clearly not in the solution area for the dotted line.
(-3,5) is clearly not in the solution area for either the solid line or the dotted line.
PTS: 2
REF: 081407a1
NAT: A.REI.12
TOP: Graphing Systems of Linear Inequalities
2. ANS: B
Strategy: Transpose the first inequality to slope intercept form (
), then input both inequalities into a
graphing calculator and eliminate wrong answers.
STEP 1.
Transpose the first inequality to slope intercept form (
).
STEP 2.
Input both inequalities into a graphing calculator and inspect the graphs.
Eliminate answer choices c and d.
STEP 3. Decide between answer choices a and b.
Eliminate answer choice a because it shows two solid lines. The graph for
Answer choice b is the correct answer.
must have a dotted line.
PTS: 2
REF: 061404a1
NAT: A.REI.12
TOP: Graphing Systems of Linear Inequalities
KEY: bimodalgraph
3. ANS: B
Strategy: Test the answer choices in the system of equations and eliminate wrong answers.
a.
c.
True
b.
True
d.
Not true
True
PTS: 2
REF: fall1301a1
NAT: A.REI.12
TOP: Graphing Systems of Linear Inequalities
4. ANS: C
Strategy: Input both inequalities into a graphing calculator and inspect the graphs.
Answer choice c is the correct answer.
PTS: 2
REF: 081506ai
NAT: A.REI.12
TOP: Graphing Systems of Linear Inequalities
5. ANS: A
Strategy: Use the slope intercept form of a line,
, to construct the inequality from the graph.
The slope is
.
The y-intercept is 4.
The equation is
.
The line is solid and the shading is above the line.
The inequality is
Answer choice a is correct.
PTS: 2
6. ANS:
a)
b)
c
REF: 061505AI
NAT: A.REI.12
TOP: Linear Inequalities
.
Oscar is wrong. The point (2,1) is not in the solution set of both inequalities.
Strategy: Use information from the graph together with the slope intercept form of a line (
) to write the
inequality
, where 2 is the slope (m) and -3 is the y-intercept b. Then, transform the new equation and
put both equations in a graphing calculator. Use the graph and the table of values to finish the system of
inequaulities on paper. Finally, determine if Oscar is right or wrong.
STEP 1. Transform
for input into a graphing calculator.
STEP 2. Input both inequalities in a graphing calculator.
STEP 3. Use the graph view and the table view to transfer the graph to paper. Be sure to make the line dotted for
. The line for
should be solid.
STEP 4. Test the point (2,1) in both equations.
True
Not True
Oscar is wrong.
PTS: 4
7. ANS:
REF: 011534a1
NAT: A.REI.12
TOP: Graphing Systems of Linear Inequalities
Strategy: Transpose the inequality, put it in a graphing calculator, then use the table and graph views to create the
graph on paper.
STEP 1. Transpose the inequality for input into a graphing calculator.
STEP 2. Inpout the inequality into a graphing calculator.
STEP 3. Use information from the graph and table views to create the graph on paper. Be sure to make the line
dotted.
PTS: 2
REF: 081526ai
NAT: A.REI.13
TOP: Graphing Linear Inequalities