# Systems of Inequalities

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10.4Systems of Inequalities

In this Math tutorial, you will learn:

• What are systems of linear inequalities?
• How do we identify the solution set (zone) of a system of linear inequalities?
• Which method is the most suitable for solving systems of inequalities?
• How do we prove the correctness of the solution when dealing with systems of inequalities?
• How do you find the minimum/maximum value of the solution set in systems of inequalities?
• How do you find the leftmost/rightmost value of the solution set in systems of inequalities?
• How do we solve systems of three linear inequalities?
• How do we solve systems of inequalities where at least one inequality is not linear?

## Introduction

Do you remember how to solve a system of linear equations? How many methods we can use for this purpose?

Check whether the following number pair are solutions for the two linear inequalities

3x - 2y > 1 and x + y ≤ 4
A(-2, 0)
B(2, 1)

Do you think there is any simpler method to check whether a number pair is a solution for two simultaneous inequalities? What does this method consist of?

## Systems of Linear Inequalities

Let's consider the two inequalities in the introduction section. Unfortunately, the first two (analytical) methods used for solving systems of linear equations (elimination and substitution method) cannot be used in systems of linear inequalities, as here we have to find a region of the coordinates plane XOY where both inequalities are true. This region can extend at different orientations, so it is impossible to identify all values that belong to the solution set without referring to a figure that shows the graphs of every single inequality.

Therefore, we can solve graphically the system of linear inequalities

3x - 2y > 1x + y ≤ 4

Obviously, as in all inequalities solved in the previous tutorial, when solving such as system only graphically, we must isolate the variable y and solve each inequality in terms of the other variable x. Then, we plot the graphs of each inequality and see where the double shaded area that shows the region where both inequalities are true extends. Therefore, only the last part of the solution reminds us of the existence of a system of linear inequalities; the rest of the solution looks like solving two different exercises containing a single inequality.

However, we can still use the first two methods when checking the solution, especially when the corresponding equations form independent systems, i.e. when they have a single common point. We can check whether this point of convergence is the same as shown in the graph or not. Hence, we can write for the two inequalities in our example:

3x - 2y > 1
-2y > -3x + 1
-2y/-2 > -3x/-2 + 1/-2
y < 3/2 x - 1/2

and

x + y ≤ 4
y ≤ -x + 4

This means the solution set includes the double shaded area below the lines y = 3x/2 -1/2 (without including the line) and y = -x + 4 (including the line itself), as shown in the figure below. Now, let's check whether points A and B in the "Introduction" section belong to the solution set or not. Let's insert points A(-2, 0) and B(2, 1) into the graph. From the graph, it is evident that point A does not belong to the solution set of the given system of linear inequalities, as it belongs only to the solution set of y ≤ -x + 4 but not to that of y < 3x/2 -1/2. On the other hand, point B belongs to the solution set of the given system of linear inequalities, as it is inside the double shaded zone.

Let's check our graphical solution by making the numerical proof. We can take either the original inequality or the revised form used to plot the graph, as the result will not change. Thus, substituting the coordinates of point A (x = -2, y = 0) in the first inequality yields

3x - 2y > 1
3 ∙ (-2) - 2 ∙ 0 > 1
-6 - 0 > 1
-6 > 1 (false)

while substituting the same point in the second inequality yields

x + y ≤ 4
-2 + 0 ≤ 4
-2 ≤ 4 (true)

Now, let's consider the coordinates of point B (x = 2, y = 1). Thus, when substituting them in the first inequality, yields

3x - 2y > 1
3 ∙ 2 - 2 ∙ 1 > 1
6 + 2 > 1
8 > 1 (true)

and when substituting the coordinates of the same point in the second inequality yields

x + y ≤ 4
2 + 1 ≤ 4
3 ≤ 4 (true)

Therefore, only point B belongs to the solution set of the given system of inequalities, as when substituting its coordinates in the system of inequalities, makes both inequalities true.

Remark! The graphing method not only makes us solve systems of linear inequalities, but it also helps us solve easier the corresponding systems of equations when the coefficients are such that they give rational numbers as solutions. Thus, instead of finding a pair of values obtained through complicated calculations, we simply plot the graphs and see what the coordinates of the intercept point are.

### Example 1

Solve the following systems of inequalities.

1. 3y - 12x ≤ 52x - 7y ≥ -1
2. x > 5 - y1/2 y - 4x ≥ -1

### Solution 1

1. First, we have to write all inequalities in the form y (?) mx + n, where (?) contains one of the four inequality symbols ( >, <, ≥ or ≤ ). It is worth to highlight the fact that we don't make this new arrangement to plot the graph line easier, as even in the actual form you can find two points that help plotting the graph. The reason why we isolate the y-variable is to identify easier the direction of the solution zone. Hence, for the first inequality, we have
3y - 12x ≤ 5
3y ≤ 12x + 5
3y/312x/3 + 5/3
y ≤ 4x + 5/3
This means the solution set of this inequality includes the zone below the graph line including this line as well.
As for the second inequality of the system, we have
2x - 7y ≥ -1
-7y ≥ -2x - 1
-7y/-7-2x/-7 - 1/-7
y ≤ 2/7 x + 1/7
This means the solution set of this inequality includes the zone below the graph line including this line as well.
Hence, the solution set will include the common zone that lies below each graph, as shown by the double shaded zone in the figure. 2. Again, we have to write all inequalities in the form y (?) mx + n, where (?) contains one of the four inequality symbols ( >, <, ≥ or ≤ ). Thus, for the first inequality, we have
x > 5 - y
x + y > 5 - y + y
x + y > 5
x + y - x > 5 - x
y > 5 - x
y> - x + 5
Thus, the solution set of this inequality alone includes the zone above the graph without the graph line.
As for the second inequality, we have
1/2 y - 4x ≥ -1
2 ∙ (1/2 y - 4x) ≥ 2 ∙ (-1)
y - 8x ≥ -2
y - 8x + 8x ≥ 8x - 2
y ≥ 8x - 2
Therefore, the solution set of this inequality alone includes the zone above the graph without the graph line. Hence, if considering the two graphs as forming an "X-letter" like shape, the solution zone of the system will include the upper quarter of the shape, as shown in the figure below. ## Determining the Minimum or Maximum Value(s) of a System of Linear Inequalities

Sometimes, we are asked to find the minimum or maximum values of a system of linear inequalities. This means finding one of the following values:

1. The maximum x-value of the system. In other words, we are asked to find the rightmost value of the solution set on the graph;
2. The minimum x-value of the system. In other words, we are asked to find the leftmost value of the solution set on the graph;
3. The maximum y-value of the system. In other words, we are asked to find the uppermost value of the solution set on the graph;
4. The minimum y-value of the system. In other words, we are asked to find the lowermost value of the solution set on the graph;
5. A combination of two minimum or maximum coordinates, where possible.

Obviously, the method used for this purpose consists of determining the solution zone by solving the given system of linear inequalities first, and eventually, solving analytically the corresponding system of linear equations to determine the intercept point of the graph with high precision - an action that is often impossible to do through the graphing method, as the solution pair may consist of rational numbers. For example, in the system of linear inequalities

2x + 3 < y3x - y ≥ 4

we arrange first the two inequalities of the system to make them appear in the form y (?) mx + n, where (?) represents one of the four inequality symbols. Thus, for the first inequality, we have

2x + 3 < y

Looking at this inequality from right to left yields

y > 2x + 3

Hence, the solution set of this inequality alone includes all points above the graph without those on the graph line.

On the other hand, for the second inequality, we have

3x - y ≥ 4
3x - y - 3x ≥ 4 - 3x
-y ≥ -3x + 4
(-1) ∙ (-y) ≥ (-1)(-3x + 4)
y ≤ 3x - 4

Therefore, the solution set of this inequality alone includes all points under the graph including those on the graph line as well.

The figure below shows this system of linear inequalities solved graphically. From the above graph, it is evident that the solution set has a minimum y-value, which is at the lowest point of the solution zone. Hence, we can find the minimum y-coordinate of the solution zone by looking at the intercept point of the two graphs. Since the graph method is not very helpful in finding a precise y-coordinate of the intersection, we solve analytically the system of the corresponding linear equations to determine this point with high precision. Thus, we have

2x + 3 = y3x - y = 4

or

2x - y = -33x - y = 4

Multiplying the first equation by (-1) to eliminate y and then adding the two equations yields

(-1) ∙ (2x - y) = (-1) ∙ (-3)3x - y = 4
-2x + y = 33x - y = 4
-2x + 3x + y - y = 3 + 4
x = 7

Now, substituting this value found for x in any of the equations (for example in the second) we are able to find the y-coordinate of the intercept, which represents the minimum value of the solution set for the original system of inequalities. Thus,

3x - y = 4
3 ∙ 7 - y = 4
21 - y = 4
y = 17

Therefore, the minimum value of the system of the inequalities

2x + 3 < y3x - y ≥ 4

is ymin = 17 without including this point, as one of the inequalities (the first) does not include it. We say, "The solution zone includes only values that have their y-coordinate more than 17".

In addition, we can find the leftmost value of the solution set by considering the x-value of the solution set. Thus, x = 7 represents the minimum x-value of the solution set, where again this value is not included, as the first inequality does not contain it.

### Example 2

Calculate the maximum or minimum x- and y-values of the solution set in the system of linear inequalities

4x - y ≤ -13x + 5 > 6y

### Solution 2

First, we have to express each inequality in the form y (?) mx + n to see whether we have to check for minimum or maximum values in the solution set. Thus, for the first inequality, we have

4x - y ≤ -1
-y ≤ -4x - 1
(-1) ∙ (-y) ≤ (-1) ∙ (-4x - 1)
y ≥ 4x + 1

and for the second inequality, we have

3x + 5 > 6y
6y < 3x + 5
6y/6 < 3x/6 + 5/6
y < x/2 + 5/6

Since the solution zone of the first inequality includes the zone above the graph (including the line) while that of the second inequality includes the zone below the graph (without the line), it is impossible to determine whether the solution zone contains a maximum or minimum without plotting the graphs. Hence, we refer to the figure below for more detailed info. From the figure, it is clear that the solution set has a maximum point at the intercept of the two graph lines made of the uppermost y-coordinate and the rightmost x-coordinate. The graphing method is not very helpful in determining these two coordinates, so we have to solve the system of the corresponding linear equations analytically. Thus, we have

4x - y = -13x + 5 = 6y
4x - y = -13x - 6y = -5

Let's express the first equation as y = 4x + 1 and solve the second equation in terms of x by substituting y with 4x + 1. Thus,

3x - 6(4x + 1) = -5
3x - 24x - 6 = -5
-21x = -5 + 6
-21x = 1
x = -1/21

As you see, this is a rational value, impossible to find directly in the graph. Hence, we say "xmax = -1/21".

As for the maximum y-value of the solution set (ymax), we substitute in the first equation the value found for x, which yields

4x - y = -1
y = 4x + 1
y = 4 ∙ (-1/21) + 1
= -4/21 + 21/21
= 17/21

Thus, the maximum value of the solution zone is ymax = 17/21, as shown in the graph.

## Solving Systems of Three Linear Inequalities

Sometimes, a third inequality is added to the normal systems of two linear inequalities, producing a system of three linear inequalities. The simplest case is when the third inequality contains a single variable and acts as a third boundary line for the solution zone. In this way, we often obtain a closed figure (triangle) as a solution zone, formed by the two-by-two intercepts of the three lines, as shown in the example below.

### Example 3

1. Identify the solution zone of the system of linear inequalities
3x + 2y < 1-2x + 5y > -2x ≥ -2
2. Check whether points A(-3, 2), B(-1, 1) and C(2, 0) belong to the solution zone of this system.
3. Find the coordinates of graphs intercepts.

### Solution 3

1. First, we have to turn all inequalities in the form y (?) mx + n to identify the individual solution zones. Thus, for the first inequality, we have
3x + 2y < 1
2y < -3x + 1
2y/2 < -3x/2 + 1/2
y < -3x/2 + 1/2
This means the solution zone of the above inequality lies under its graph line.
For the second inequality, we have
-2x + 5y > -2
5y > 2x - 2
5y/5 > 2x/5 - 2/5
y > 2x/5 - 2/5
This means the solution zone of the above inequality lies above its graph line.
As for the third inequality, it is not necessary to do any operation as it is clear that the solution zone lies on the right of the vertical line x = 2, including the line itself. Therefore, based on the above findings we obtain the following figure: 2. Now, we have to check whether points A(-3, 2), B(-1, 1) and C(2, 0) belong to the solution zone of this system by substituting each point in the inequalities of the system and see whether this substitution gives true inequalities or not. Thus, for point A(x = -3 and y = 2), we have
3x + 2y < 1-2x + 5y > -2x ≥ -2
3 ∙ (-3) + 2 ∙ 2 < 1-2 ∙ (-3) + 5 ∙ 2 > -2-3 ≥ -2
-9 + 4 < 1-6 + 10 > -2-3 ≥ -2
-5 < 1 (true)4 > -2 (true)-3 ≥ -2 (false)
Since not all inequalities are true for the coordinates of point A, this point does not belong to the solution set of this system of inequalities.
For point B(-1, 1) where x = -1 and y = 1, we obtain
3x + 2y < 1-2x + 5y > -2x ≥ -2
3 ∙ (-1) + 2 ∙ 1 < 1-2 ∙ (-1) + 5 ∙ 1 > -2-1 ≥ -2
-3 + 2 < 12 + 5 > -2-1 ≥ -2
-1 < 1 (true)7 > -2 (true)-1 ≥ -2 (true)
Since all inequalities are true for the coordinates of point B, this point belongs to the solution set of this system of inequalities.
Last, let's check whether point C(2, 0) where x = 2 and y = 0 belongs to the solution zone of the given system of linear inequalities or not. We have
3x + 2y < 1-2x + 5y > -2x ≥ -2
3 ∙ 2 + 2 ∙ 0 < 1-2 ∙ 2 + 5 ∙ 0 > -22 ≥ -2
6 + 0 < 1-4 + 0 > -22 ≥ -2
6 < 1 (false)-4 > -2 (false)2 ≥ -2 (true)
Since not all inequalities are true for the coordinates of point C, this point does not belong to the solution set of this system of inequalities.
3. To find the intercepts, i.e. the points that form the vertices of the triangle which includes the solution zone, we consider two by two the corresponding equations obtained by the three given inequalities. Thus, for the first vertex, we obtain the following system of linear equations
3x + 2y = 1-2x + 5y = -2
Let's eliminate the variable x by multiplying the first inequality by 2 and the second one by 3 and eventually add the two equations. Thus,
2 ∙ (3x + 2y) = 2 ∙ 13 ∙ (-2x + 5y) = 3 ∙ (-2)
6x + 4y = 2-6x + 15y = -6
-19y = -4
y = -4/19
We find the x-coordinate of the first vertex by substituting the above y-coordinate in any of the equations in the system; for example in the first. Thus,
3x + 2 ∙ (-4)/19 = 1
3x - 8/19 = 1
3x = 1 + 8/19
3x = 19/19 + 8/19
3x = 27/19
x = 27 ÷ 3/19
= 9/19
Therefore, one of the intercepts (the bottom left) has the coordinates (9/19, -4/19).
Now, let's find the second intercept by solving the system
3x + 2y = 1x = -2
This system is solved easier by substituting x = -2 in the first equation. This yield,
3 ∙ (-2) + 2y = 1
-6 + 2y = 1
2y = 1 - (-6)
2y = 7
y = 7/2
Therefore, the second intercept (the top-left one) is at (-2, 7/2).
Last, we find the third intercept by solving the linear system
-2x + 5y = -2x = -2
Again, we solve this system by substituting x = -2 in the first equation. Thus,
-2 ∙ (-2) + 5y = -2
4 + 5y = -2
5y = -2 - 4
5y = -6
y = -6/5
Therefore, the third graphs' intercept (the bottom-left one) has the coordinates (-2, -6/5).

It is not always possible to obtain a closed region (like the triangle in the above example) as a solution zone for a system of three linear inequalities. For example, if two of the linear inequalities are dependent (i.e. when they have parallel graphs), we obtain a zone that is limited in two or three directions but unlimited in the fourth, as the graphs form a figure where two parallel lines are intercepted by a third one. Let's consider an example to clarify this point.

### Example 4

Solve graphically the system of linear inequalities

1. x - 2y < 13x - 6y ≤ 5x < 2
2. Find the coordinates of the highest point of the solution set.

### Solution 4

1. We have two write the first inequalities in the form y (?) mx + n. Thus, for the first inequality, we have
x - 2y < 1
-2y < 1 - x
-2y/-2 < 1 - x/-2
y > -1/2 + -x/-2
y > x/2 - 1/2
Hence, the solution set of this inequality alone consists of the region above the graph.
For the second inequality, we have
3x - 6y ≤ 5
-6y ≤ -3x + 5
-6y - 6 ≤ -3x/-6 + 5/-6
y ≥ x/2 - 5/6
Hence, the solution set of this inequality alone consists of the region above the graph including the graph line.
The third inequality includes the region on the left of the graph without including the vertical line x = 2/.
Hence, the graphical solution to this system of inequalities is ## Solving Systems of Inequalities where one inequality is Quadratic and the other is Linear

When solving systems of inequalities where not all are linear (for example, when one inequality is quadratic), we use the same approach as in systems of linear inequalities. The only difference is that one of the lines is a parabola, so the solution zone has one curved side.

Another thing to point out in such systems is that if we are interested to find any minimum or maximum point as in the previous examples, we must solve the corresponding system of equations only by substitution method. Thus, one of the variables in the linear equation is expressed in terms of the other variable and we substitute it in the quadratic equation. Let's consider an example to clarify this point, where theory and exercise are combined to have a better understanding.

### Example 5

Solve the following system of inequalities.

3x - 2y > 1y ≥ x2-2x + 1

### Solution 5

First, we identify the direction of the solution set for the linear inequality. We do this by writing the first inequality in the form y (?) mx + n. Thus,

3x - 2y > 1
-2y> - 3x + 1
-2y - 2 > -3x/-2 + 1/-2
y < 3x/2 - 1/2

Hence, the solution zone of this inequality alone lies under the graph's line y = 3x/2 - 1/2.

On the other hand, the quadratic equation is already in its regular form y (?) ax2 + bx + c, where a = 1, b = -2 and c = 1, while the inequality symbol which replaces the question mark is " ". This means the solution zone lies above the parabola's graph.

The best thing to do is to solve the system of the corresponding equations. The number of solutions depends on the sign of the discriminant obtained when the linear equation is substituted into the quadratic one. In the specific case, we have

3x - 2y = 1y = x2 - 2x + 1

We can write 3x - 1 = 2y, multiply the quadratic equation by 2 and then express y in the quadratic equation in terms of x. In this way, we obtain

2y = 2x2 - 4x + 2
3x - 1 = 2x2-4x + 2
2x2 - 4x + 2 - 3x + 1 = 0
2x2 - 7x + 3 = 0

This is a second-order equation with one variable, where its roots represent the intercept of the two original graphs. We have a = 2, b = -7 and c = 3. Thus, the discriminant Δ is

∆ = b2 - 4ac
= (-7)2 - 4 ∙ 2 ∙ 3
= 49 - 24
= 25

Since the discriminant is positive, we have two intercepts for our graphs. Their x-coordinates are found by solving the second-order equation above. Expressing the two intercepts by A and B respectively yields

xA = -b - √∆/2a
= -(-7) - √25/2 ∙ 2
= 7 - 5/4
= 2/4
= 1/2
= 0.5

and

xB = -b + √∆/2a
= -(-7) + √25/2 ∙ 2
= 7 + 5/4
= 12/4
= 3

The corresponding y-values for points A and B are

yA = xA2 - 2xA + 1
= 0.52 - 2 ∙ 0.5 + 1
= 0.25 - 1 + 1
= 2.25
yB = xB2 - 2xB + 1
= 32 - 2 ∙ 3 + 1
= 9 - 6 + 1
= 4

Hence, the two intercepts are A(0.5, 2.25) and B(3, 4).

Another important point of the solution set that helps identify (in this case) the lowest point of the solution set is the parabola vertex. We have briefly discussed this point in previous tutorials, where the formulas used to find the vertex V of a parabola are

xV = -b/2a and yV = -∆/4a

Substituting the known values for the original quadratic equation (a = 1, b = -2 and c = 1) yields

xV = -(-2)/2 ∙ 1
= 2/2
= 1

and

yV = -(b2 - 4ac)/4a
= -(-2)2 - 4 ∙ 1 ∙ 1/4 ∙ 1
= 4 - 4/4
= 0

Hence, point V(1, 0) is the vertex of the parabola, which at the same time acts as the lowest point of the solution set for the original system of inequalities.

The graph solution for the system is shown below. We can draw the following conclusions by looking at the graph:

1. Vertex V is the minimum point of the system's solution set.
2. Point A is the leftmost point of the solution set but point A itself is not included in this set.
3. Point B is both the maximum and the rightmost value of the solution set, where it is also excluded from the solution values.
4. The two intercepts A and B can be used to plot the linear graph, since plotting the graph of a straight line requires only two known points.

We can use the same approach for systems with more than two inequalities, where not all of them are linear.

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