Sign In
| 8 Theory slides |
| 10 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
Here are a few recommended readings before getting started with this lesson.
An absolute value inequality in two variables is an inequality that contains an absolute value expression and shows the relationship between two variables. If the relationship is linear, then the inequality is similar to the equation of an absolute value function. Example [-0.8em] Equation: y = |8x-3|+2 Inequality: y ≤ |8x-3|+2 While the graph of an absolute value equation is a V-shaped graph, the graph of a two-variable absolute value inequality is a region. When graphing a two-variable absolute value inequality, its boundary line plays an important role.
The boundary line of the inequality can be determined by replacing the inequality symbol with the equals sign. Inequality: & y-3 > |3x-6| Boundary Line: & y-3 = |3x-6| The graph of the boundary line can be drawn using a table of values. To make calculations easy, first y will be isolated. y-3 = |3x-6| ⇕ y=|3x-6|+3 Now some points on the boundary line can be found.
x | y=|3x-6|+3 | y |
---|---|---|
0 | y=|3( 0)-6|+3 | 9 |
1 | y=|3( 1)-6|+3 | 6 |
2 | y=|3( 2)-6|+3 | 3 |
3 | y=|3( 3)-6|+3 | 6 |
4 | y=|3( 4)-6|+3 | 9 |
Plot the points and draw the boundary line. Since the given inequality is strict, the boundary line will be dashed.
x= 0, y= 0
Zero Property of Multiplication
Subtract terms
|-6|=6
The test point is not a solution, so the region that does not contain the test point — inside the V-shaped graph — represents the solution set for the inequality. Finally, the appropriate region can be shaded.
Analyze the graph of the given absolute value inequality in two variables to determine the proper inequality symbol.
Dylan writes an absolute value equation in two variables for the shape of a mirror. y-4 = - |5/2x+5| He knows that only the outside of the mirror is reflective.
LHS+4=RHS+4
a = 2* a/2
Factor out 5/2
|a* b|= |a| * |b|
|5/2|=5/2
a=- (- a)
x | - 5/2|x+2| + 4 | y = -5/2 |x+2 | + 4 |
---|---|---|
- 4 | - 5/2| - 4+2| + 4 = - 1 | - 1 |
- 3 | - 5/2| - 3+2| + 4 = 1.5 | 1.5 |
- 2 | - 5/2| - 2+2| + 4 = 4 | 4 |
- 1 | - 5/2| - 1+2| + 4 = 1.5 | 1.5 |
0 | - 5/2| 0+2| + 4 = - 1 | - 1 |
Finally, plot the points and draw the absolute value equation.
As can be seen, the mirror has a V-shape with a vertex of (- 2,4) in the coordinate plane.
The shaded region represents the y-values greater than - | 52x +5 |+4. The inequality symbol will be strict because the graph of the mirror is not part of the solution set. Inequality: y > - |5/2x+5 | + 4 In this example, the mirror represents the boundary line of the inequality, the spherical light source represents a test point that satisfies the inequality, and the shaded region represents the graph of the inequality.
Ignacio buys an antique object at an auction. He plans to sell it in a few years. He describes the value of the object y (in thousands dollars) after x years as follows. y = 0.4 |x-4| + 3 He is also willing to sell the object at any time for any price greater than or equal to the value found by the above equation.
0.4|x-4|+3 If Ignacio can find someone who is ready to pay greater than or equal to the amount after x years, the object is sold. Therefore, the following inequality will describe the situation. y ≥ 0.4|x-4|+3 To graph this absolute value inequality, its boundary line will be drawn first. To do so, make a table of values for y=0.4|x-4|+3. Note that since x represents the number of years, it cannot be negative.
x | y=0.4|x-4|+3 | y |
---|---|---|
0 | y=0.4| 0-4|+3 | 4.6 |
2 | y=0.4| 2-4|+3 | 3.8 |
4 | y=0.4| 4-4|+3 | 3 |
6 | y=0.4| 6-4|+3 | 3.8 |
8 | y=0.4| 8-4|+3 | 4.6 |
Plot the points and draw the boundary line. Since the inequality is non-strict, the boundary line will be solid. Note that only positive values of x and y makes sense in the context of the situation.
In the same auction, Emily was also interested in the object that Ignacio bought. Emily knows that Ignacio will sell it sooner or later. The maximum amount of money Emily can allocate to buy Ignacio's object is represented by the following graph.
To write the equation of the boundary line, use the vertex form of an absolute value equation y=a|x-h|+k, where (h,k) is the vertex of the absolute value equation.
On the given graph, x represents the years after the auction and y represents the price of the object in thousands of dollars. Since Emily can allocate any amount less than $3800 for the object 4 years after the auction, these values correspond to point (4,3.8), which is the vertex of the boundary line.
x= 10, y= 5
Subtract term
|6|=6
LHS-3.8=RHS-3.8
.LHS /6.=.RHS /6.
Rearrange equation
less than,<. Boundary Line: & y = 0.2|x-4|+3.8 Inequality:& y < 0.2|x-4|+3.8
When absolute value inequalities in two variables are drawn on the same coordinate plane, they might intersect. The region where the graphs of the inequalities intersect represents the solutions that satisfy both inequalities. Consider the following absolute value inequalities. Inequality I: & y ≥ 0.4|x-4|+3 Inequality II: & y < 0.2|x-4|+3.8 There is a small region where the graphs of these inequalities intersect.
Any point in this region is the solution to both inequalities. In the context of the last two examples, Ignacio and Emily can make a deal as long as the values stay in that region.
Consider the absolute value inequality. y ≤ 0.5|x-3|+1 Which graph represents the solution set of the inequality?
To graph an inequality, remember that, we always begin by determining the boundary line. The boundary line can be written by replacing the inequality sign with an equal sign. &Inequality &&Boundary Line & y ≤ 0.5|x-3|+1 && y = 0.5|x-3|+1 We have an absolute value equation for the boundary line. Notice that it is in the vertex form. Vertex Form [-1em] y= a|x- h|+ k ⇓ y= 0.5|x- 3|+ 1 The vertex is ( 3, 1). To graph the boundary line, we can make a table of values.
x | 0.5|x-3|+1 | y = 0.5|x-3|+1 |
---|---|---|
1 | 0.5| 1-3|+1 | 2 |
2 | 0.5| 2-3|+1 | 1.5 |
3 | 0.5| 3-3|+1 | 1 |
4 | 0.5| 4-3|+1 | 1.5 |
5 | 0.5| 5-3|+1 | 2 |
Let's plot the points ad draw the boundary line. Since the inequality is non-strict, the boundary line will be solid.
We will test a point to decide which region we should shade. Let the point be (0,0). If it satisfies the inequality, we will shade the region that contains the point. Otherwise, we will shade the other region.
Since the point satisfies the inequality we will shade the region that contains the point.
Therefore, the correct option is A.
Which graph represents the solution set of the inequality |3x-2y| ≥ - 8 ?
We know that all absolute value expressions are greater than or equal to zero. |3x-2y| ≥ 0 It follows then that all absolute value expressions are greater than or equal to any negative number. |3x-2y|≥ 0 ⇒ |3x-2y| ≥ - 8 This means all values of x and y will satisfy the given inequality. Therefore the solution of this inequality is all real numbers and we should shade the whole coordinate plane.
The answer is D.
Suppose that we were given an absolute value inequality, where the absolute value expression is less than a negative number. Let's change the inequality symbol of the given inequality.
|3x-2y| < - 8
Again, since all absolute value expressions are greater than or equal to zero, we cannot find any x- and y-values that satisfy the inequality |3x-2y| < - 8. Therefore, the solution set of such an inequality is the empty set.
We can use a test point to determine which region of the plane represents the solution set of an inequality. However, if the test point lies on the boundary line, it will give us no new information!
This is why, we prefer to use a point that is not on the boundary line. In this case, we need to determine inequalities whose boundary line passes through the origin. Let's first determine the boundary lines.
Inequality | Boundary Line | |
---|---|---|
I | y > |x| | y=|x| |
II | y ≥ |2x| | y=|2x| |
III | y + 2 < |x| | y+2= |x| |
IV | y+2 > |x-2| | y+2=|x-2| |
V | y > - 2|x-2|-2 | y = - 2|x-2|-2 |
We will now substitute (0,0) into the equations of the boundary lines. If the substitution produce a true statement, it means that the boundary line passes through the origin.
Boundary Line | Substitute (0,0) | Is it a true statement? | |
---|---|---|---|
I | y=|x| | 0=| 0| | Yes |
II | y=|2x| | 0=|2* 0| | Yes |
III | y+2= |x| | 0+2=| 0| | No |
IV | y+2=|x-2| | 0+2=| 0-2| | Yes |
V | y = - 2|x-2|-2 | 0=- 2| 0-2|-2 | No |
For inequalities I, II, and IV, the point (0,0) lies on the boundary line. Therefore, we will not be able to use it to determine the regions representing the solution sets of the inequalities.
To determine the shape of the region, we should first draw each inequality separately. Then we will combine the graphs on the same coordinate plane and find their intersection. Let's start!
To graph |y|≤ 3, we will have to create a compound inequality first. Since |y| is less than or equal to x, the word and
will be used to form the compound inequality.
|y|≤ 3 ⇔ y≤ 3 and y≥ - 3
We can determine the boundary lines of this compound inequality by changing the inequality symbols to equal signs.
Boundary Lines
y=3 and y=- 3
Both boundary lines are horizontal. Since the inequalities are non-strict, the lines will be solid. The solution set of this inequality contains all coordinate pairs whose y-value is less than or equal to 3 and greater than or equal to - 3. This means that we should shade the region between the lines.
Also, notice that the boundary lines are parallel.
To determine the boundary line of the second inequality, we need to exchange the inequality symbol for an equal sign. Inequality & Boundary Line y ≤ - 2|x+3|+6 & y = - 2|x+3|+6 We see that the boundary line equation is an absolute value equation in vertex form. Vertex Form [-1em] y= a|x- h|+ k ⇓ y= - 2|x-( - 3)|+ 6 The vertex is ( - 3, 6) and its graph is symmetric about the line x= - 3. To graph the boundary line, we can make a table of values.
x | - 2|x+3|+6 | y = - 2|x+3|+6 |
---|---|---|
- 6 | - 2| - 6+3|+6 | 0 |
- 3 | - 2| - 3+3|+6 | 6 |
0 | - 2| 0+3|+6 | 0 |
Let's plot (- 6,0), (- 3,6), and (0,0) and draw the boundary line. Remember that the boundary line is solid since the inequality is non-strict.
Next, we decide which side of the boundary line we should shade. We can do this by testing a point that does not lie on the boundary line. If the point satisfies the inequality, it lies in the solution set. If not, we will shade the other region. Let's use ( - 3, 3).
Because (- 3,3) created a true statement, we will shade the region that contains this point.
Let's draw the graphs of the inequalities on the same coordinate plane.
Finally, we can view only the overlapping region.
We know that the boundary lines of the first inequality are parallel. Since the overlapping region is a quadrilateral with one pair of parallel sides, it is a trapezoid.