System of Inequalities
A system of inequalities is a set of two or more inequalities in one or more variables. Systems of inequalities are used when a problem requires a range of solutions, and there is more than one constraint on those solutions.
Leon is the manager of a textile factory. His workers have a total of 400 man-hours this week for him to allocate. Every 10 shirts will take 2 hours to produce. Every 10 pairs of pants will take 3 hours to produce. He is required to produce at least 30 shirts and at least 20 pairs of pants this week. How can Leon allocate his man-hours, and how will this affect his production?
This problem is ideal to be modeled with a system of inequalities. There are multiple solutions, and there is more than one constraint on those solutions.
Let \(s\) be the number of shirts produced, and let \(p\) be the number of pairs of pants produced. The number of hours producing shirts is \(\frac{2}{10}s=\frac{1}{5}s,\) and the number of hours producing pants is \(\frac{3}{10}p.\) Leon's production is constrained by the following system of inequalities:
\[\begin{cases} \begin{array}{rl} \frac{1}{5}s+\frac{3}{10}p &\le 400 \\ s &\ge 30 \\ p &\ge 20.\ _\square \end{array} \end{cases}\]
As with the example above, systems of inequalities are often used to define the constraints on a solution. When a problem requires you to pick an optimal solution, then this requires linear programming or other optimization techniques (combinatorial optimization or optimization with calculus).
Contents
Definitions and Notation
Systems of inequalities follow much of the same notation as linear inequalities.
\(>\) is the greater than symbol. The quantity to the left of the symbol is greater than the quantity to the right.
\(<\) is the less than symbol. The quantity to the left of the symbol is less than the quantity to the right.
\(\ge\) is the greater than or equal to symbol. The quantity to the left of the symbol is either greater than the quantity to the right, or it is equal to the quantity to the right.
\(\le\) is the greater than or equal to symbol. The quantity to the left of the symbol is either less than the quantity to the right, or it is equal to the quantity to the right.
\(\ne\) is the not equal to symbol. The quantity to the left of the symbol is not equal to the quantity to the right.
Inequality notations are read left to right, as in \(2 > 1\), two is greater than one.
Systems of inequalities use some notation from sets.
An intersection of two inequalities contains the numbers that satisfy both inequalities. It is denoted with the \(\cap\) symbol between the two inequalities:
\[x>-2\ \cap\ x \le 5.\]
In the case of intersections that give values for the variable that are between two numbers, \(x\) can be written between the inequality symbols. The following example is equivalent to the example above:
\[-2 < x \le 5.\]
An intersection of inequalities is also sometimes denoted with a left curly bracket, \(\{\), placed to the left of the inequalities. This is more common when the system of inequalities is in two or more variables:
\[\begin{cases} y<2x-3 \\ y\ge 3. \end{cases} \]
A union of two inequalities contains the numbers that satisfy the first or the second inequality. It is denoted with the \(\cup\) symbol between the two inequalities:
\[x<0\ \cup\ x>6.\]
The solution set of a system of inequalities is often written in set builder notation:
\[\{x \mid x<0\ \cup\ x>6\},\]
which reads "The set of all \(x\) such that \(x\) is less than 0 or \(x\) is greater than 6."
Systems of inequalities can also be denoted with interval notation.
Interval notation is a way to describe continuous sets of real numbers by the numbers that bound them. Intervals, when written, look somewhat like ordered pairs. However, they are not meant to denote a specific point. Rather, they are meant to be a shorthand way to write an inequality or system of inequalities.
Intervals are written with rectangular brackets or parentheses, and two numbers delimited with a comma. The number on the left denotes the least element or lower bound. The number on the right denotes the greatest element or upper bound.
The rectangular bracket symbols, \([\ ],\) are used to describe sets with a "less than or equal to" or a "greater than or equal to" element, respectively. They correspond to the \(\ge\) and \(\le\) symbols:
\[\begin{array}{lc} \text{Inequality:} & 3 \le x \le 9 \\ \text{Interval:} & [3,9]. \end{array}\]
In this case, \(x\) could equal \(3\) or \(9\).
The parentheses symbols, \( (\ ), \) are used to describe sets with a lower bound or upper bound, respectively. They correspond to the \(>\) and \(<\) symbols:
\[\begin{array}{lc} \text{Inequality:} & -1 < x < 4 \\ \text{Interval:} & (-1,4). \end{array}\]
In this case, \(x\) does not equal \(-1\) or \(4\).
The different types of brackets can be used in the same interval:
\[\begin{array}{lc} \text{Inequality:} & -3 \le x < 5 \\ \text{Interval:} & [-3,5) \end{array}\]
If an interval has no lower bound or upper bound, then the \(-\infty\) or \(\infty\) symbols are used. These symbols are always used with a parentheses bracket, because infinity is not a number that can be included in a set:
\[\begin{array}{lc} \text{Inequality:} & x \le 7 \\ \text{Interval:} & (-\infty,7] \\\\ \text{Inequality:} & x >2 \\ \text{Interval:} & (2,\infty). \end{array}\]
Intersections and unions of intervals can be written with the \(\cap\) or \(\cup\) symbols:
\[\begin{array}{lc} \text{Inequality:} & x \le -4\ \cup\ 0 < x < 8 \\ \text{Interval:} & (-\infty,-4]\ \cup\ (0,8)\\\\ \text{Inequality:} & x \ne 1\\ \text{Intervals:} & (-\infty,1)\ \cup\ (1,\infty). \end{array}\]
Systems of Inequalities - One Variable
Strategy for Solving Systems of Inequalities in One Variable
Solve each inequality individually. Find the intervals of solutions.
Simplify the intersection or union of intervals in the system. A number line graph can be used to help visualize the intervals.
A carnival has three rides. The swing ride requires riders to be at least 90 cm tall. The roller coaster requires riders to be at least 120 cm tall. The merry-go-round requires riders to be between 70 cm and 130 cm tall (inclusive).
Find the intervals of heights (in cm) that will allow visitors of the carnival to
- ride all rides
- ride at least one ride.
Let \(h\) be the height (in cm) of a visitor to the carnival. The required heights of each ride can be described with a system of inequalities:
\[\begin{cases} \begin{align} h &\ge 90 \\ h &\ge 120 \\ 70 \le h &\le 130. \end{align} \end{cases}\]
If a visitor to the park would want to ride all the rides, then his/her height would have to satisfy all of these inequalities. This would require the intersection of the heights:
\[\begin{array}{lc} \text{Inequality:} & 120 \le h \le 130 \\ \text{Interval:} & [120,130]. \end{array}\]
Closed points represent that \(h\) includes the points \(120\) and \(130\); open points (open circles) would be used to represent that these points are excluded, and arrows stretching off the graph would represent that the number stretches to infinity.If a visitor to the park would want to ride at least one of the rides, then his/her height would have to satisfy one or more of the inequalities. This would require the union of the inequalities:
\[\begin{array}{lc} \text{Inequality:} & 70 \le h \le 130 \\ \text{Interval:} & [70,130].\ _\square \end{array}\]
A system of inequalities is often used to describe the solutions of polynomial inequalities, absolute value inequalities, and rational inequalities.
Main Article: Polynomial Inequalities
Solve the inequality
\[x^2+2x>15.\]
Subtract 15 from both sides of the inequality:
\[x^2+2x-15>0.\]
Factor the quadratic:
\[(x+5)(x-3) >0.\]
The left side of the inequality will be positive if both factors are positive, or if both factors are negative. This occurs when \(x>3\) or when \(x<-5.\) The solution is a union:
\[\begin{array}{lc} \text{Inequality:} & x<-5\ \cup\ x>3 \\ \text{Intervals:} & (-\infty,-5)\ \cup\ (3,\infty).\ _\square \end{array}\]
Main Article: Absolute Value Inequalities
Solve the inequality
\[|x+3|+|x-1|\ge 4.\]
Case 1. \((x+3)\) and \((x-1)\) are both negative. The absolute value operations will negate these expressions:
\[\begin{align} -(x+3)-(x-1) \ge 4 \\ -2x \ge 6 \\ x \le -3. \end{align}\]
Case 2. \((x-1)\) is negative and \((x+3)\) is positive. The absolute value operations will do nothing to the \((x+3)\) expression and negate the \((x-1)\) expression:
\[\begin{align} (x+3)-(x-1) &\ge 4 \\ 2 &\ge 4. \end{align}\]
This is impossible, so there are no solutions for this case.
Case 3. \((x+3)\) and \((x-1)\) are both positive. The absolute value operations will not do anything to either expression:
\[\begin{align} (x+3)+(x-1) &\ge 4 \\ 2x &\ge 2 \\ x &\ge 1. \end{align}\]
The solution is a union of inequalities:
\[\begin{array}{lc} \text{Inequality:} & x\le -3 \ \cup\ x\ge 1 \\ \text{Intervals:} & (-\infty,-3] \ \cup\ [1,\infty).\ _\square \end{array}\]
Main Article: Rational Inequalities
Solve the inequality
\[\frac{x^2-x-9}{x-1}\le 1.\]
Case 1. \(x-1>0.\) Multiplying both sides of the inequality by \(x-1\) will not change the inequality sign:
\[\begin{align} x^2-x-9 &\le x-1 \\ x^2-2x-8 &\le 0 \\ (x-4)(x+2) &\le 0. \end{align}\]
One of the factors must be positive while the other is negative. Since \(x-4<x+2,\) the \((x-4)\) factor must be negative while the \((x+2)\) is positive. This can only occur when \(-2\le x \le 4.\) However, this is subject to the condition on this case that \(x>1.\) Simplifying the intersection gives
\[\begin{array}{lc} \text{Inequality:} & 1<x\le 4 \\ \text{Interval:} & (1,4]. \end{array}\]
Case 2. \(x-1<0.\) Multiplying both sides of the inequality by \(x-1\) will reverse the inequality sign:
\[\begin{align} x^2-x-9 &\ge x-1 \\ x^2-2x-8 &\ge 0 \\ (x-4)(x+2) &\ge 0. \end{align}\]
The left side of the inequality will be greater than or equal to 0 when both factors are positive, both factors are negative, or one of the factors is zero. This can only occur when \(x\le -2\ \cup\ x\ge 4.\) However, this is subject to the condition on this case that \(x<1.\) Simplifying the intersection gives
\[\begin{array}{lc} \text{Inequality:} & x\le -2 \\ \text{Interval:} & (-\infty,-2]. \end{array}\]
Then, the solution for both cases is
\[\begin{array}{lc} \text{Inequality:} & x\le -2\ \cup\ 1<x\le 4 \\ \text{Interval:} & (-\infty,-2]\ \cup\ (1,4].\ _\square \end{array}\]
Systems of Linear Inequalities in Two Variables
The solution to a system of inequalities in two variables is often shown as a shaded graph on the coordinate plane. Shaded regions show the areas that contain points in the solution. If a line is solid, then the points on the line are contained in the solution. If a line is dashed, then the points on the line are not contained in the solution, but any adjacent shaded region does contain points in the solution.
\[\begin{cases} y \le 2x+3 \\ y > -\frac{1}{3}x-1 \end{cases}\]
The shaded region is the intersection of the inequalities. Every point in the shaded region and on the solid ray is part of the solution. The dashed lines and rays are not part of the solution.
The following process works by isolating the \(y\) variable in each inequality. Since greater \(y\) values are higher up in the coordinate plane, a \(>\) or \(\ge\) symbol means that the solution exists above the line of the inequality. Likewise, a \(<\) or \(\le\) symbol means that the solution exists below the line of the inequality.
Graphing Linear Systems of Inequalities: Slope-Intercept Shading Method
Put each inequality into slope-intercept form.
Graph the line that bounds each inequality. If the symbol is \(\le\) or \(\ge,\) then the line should be solid to show that the points on the line are included in the solution. If the symbol is \(<,\) \(>,\) or \(\ne,\) then the line should be dashed to show that the points on the line are not included in the solution.
For each inequality, if the symbol is \(\ge\) or \(>,\) then shade above the line. If the symbol is \(\le\) or \(<,\) then shade below the line. If the symbol is \(\ne,\) then shade on both sides of the line.
If the system is a union, then your graph is complete. If the system is an intersection, then only the regions that are a part of all inequalities are in the solution. You must erase all shading, lines, and rays that are not in the solution.
Graph the union of inequalities
\[\begin{array}{ccc} y>3x & \cup & y<2x. \end{array}\]
Begin by graphing the line of each inequality. The symbols are \(>\) and \(<,\) so the lines should be dashed.
The first inequality has \(y>3x,\) so the shading should be above the line.
The second inequality has \(y<2x,\) so the shading should be below the line.
Since this system is a union, all shaded parts are part of the inequality. For clarity, each shaded region should be the same color, and the dashed lines in the shaded region should be made solid to indicate that they are part of the solution.
Every point in the shaded region has either \(y>3x\) or \(y<2x.\) \(_\square\)
Latoya manages a factory that produces ready-to-assemble furniture. She is planning how to allocate her resources to her equipment for producing desks. Machine \(\text{A}\) can produce 6 desks per hour, and costs $100 for every hour that it runs. Machine \(\text{B}\) can produce 10 desks per hour, and costs $200 for every hour that it runs. Latoya has available workers that can run the machines for up to 50 hours this week, and she has allocated $8000 of her budget to run these machines. Make a graph that shows how she can allocate her resources to produce desks.
Let \(a\) be the number of hours that machine \(\text{A}\) runs, and let \(\text{B}\) be the number of hours that machine \(\text{B}\) runs. A system of inequalities can be written to describe the constraints on how these machines are used.
First, describe how the machines are constrained by time:
\[a+b \le 50.\]
Then, describe how the machines are constrained by cost:
\[\begin{align} 100a+200b &\le 8000 \\ a+2b &\le 80. \end{align}\]
In addition, it is not possible for the machines to run less than 0 hours:
\[\begin{align} a &\ge 0 \\ b &\ge 0. \end{align}\]
Let \(a\) be graphed on the \(x\)-axis and let \(b\) be graphed on the \(y\)-axis. Solving for \(b\) in each inequality except for \(a\ge 0\) gives the system
\[\begin{cases} b \le -a+ 50 \\ b \le -\frac{1}{2}a+40 \\ a \ge 0 \\ b \ge 0. \end{cases}\]
Graph each of the lines.
Since the first inequality has a \(<\) symbol, shade below the line.
Apply the same principle to shade all the other inequalities. The \(a \ge 0\) inequality is shaded to the right, because greater values of \(a\) exist further right on the graph.
This leaves the entire graph shaded. However, this system is an intersection, since all of the inequalities must be satisfied. Therefore, all shading that is not in all inequalities must be erased.
The shaded region that remains is the solution to the system of inequalities. Any ordered pair in this solution would give Latoya a feasible way to plan the use of her equipment. You might have noticed that the number of desks produced was not included in this analysis. If the goal is to develop an optimal way to produce desks, then one must use linear programming. \(_\square\)
The following process works because the lines of each inequality partition the coordinate plane into regions. If a point in the coordinate plane satisfies the system, then all points in the same region as that point will also satisfy the system.
Solving Linear Systems of Inequalities: Test Point Method
Graph the line for each inequality. Follow the same convention for dashed and solid lines as before.
For each region that the lines partition, select a point in that region. Test the point by substituting the \(x\) and \(y\) values into each inequality.
If the system is a union, then the test point must satisfy just one of the inequalities. If the system is an intersection, then it must satisfy all inequalities. If the test point satisfies the system, then shade the region that the test point is in.
Erase all lines and rays that are not in the solution.
Graph the solution to the following system of inequalities
\[\begin{cases}\begin{align} 2x+3y &\le 4 \\ x-4y &> 2. \end{align}\end{cases}\]
Note that this system is an intersection. First, graph the line for each inequality. The first inequality should get a solid line, and the second inequality should get a dashed line. These lines partition the graph into 4 regions.
Select a point from each region and test it on the inequalities:
\[\begin{array}{ccl} \boxed{1}: & (0,0) & \begin{cases}\begin{array}{rlc} 2(0)+3(0) & \le 4 & \checkmark \\ 0-4(0) & > 2 & \text{x} \end{array}\end{cases} \\ \boxed{2}: & (2,1) & \begin{cases}\begin{array}{rlc} 2(2)+3(1) & \le 4 & \text{x} \\ 2-4(1) & > 2 & \text{x} \end{array}\end{cases} \\ \boxed{3}: & (3,0) & \begin{cases}\begin{array}{rlc} 2(3)+3(0) & \le 4 & \text{x} \\ 3-4(0) & > 2 & \checkmark \end{array}\end{cases} \\ \boxed{4}: & (2,-1) & \begin{cases}\begin{array}{rlc} 2(2)+3(-1) & \le 4 & \checkmark \\ 2-4(-1) & > 2 & \checkmark \end{array}\end{cases} \end{array}\]
The only point that satisfies both inequalities is the point in region 4. Shade this region, and remove all solid rays that are not in this region.
\(_\square\)
What are the relative weights of these shapes?
Systems of Non-linear Inequalities in Two Variables
Many of the same principles that apply to linear systems also apply to non-linear systems. If the \(y\) variable can be isolated on one side of an inequality, then the shade up/down method can be used. Otherwise, the test point method can be used.
A dog is chained to a post on a fence. The length of the chain is 20 m. The dog can roam out into the yard as far as the chain allows, but the fence extends further than 20 m on either side of the post.
Let the fence be located on the \(x\)-axis, let the post be located on the origin, and let the yard be located in the \(1^\text{st}\) and \(2^\text{nd}\) quadrants. Write and graph a system of inequalities that describes the possible locations that the dog could visit.
Since the fence is located on the \(x\)-axis, and the dog can only move up in the positive \(y\) direction from it, this gives the inequality
\[y>0.\]
The \(>\) symbol is used because the dog is blocked by the fence.
The dog is constrained by the chain, which can extend out to a circle with radius 20. The equation of the circle is \(x^2+y^2=400.\) To show that the dog can move to any point up to the radius of this circle, this equation is converted to an inequality:
\[x^2+y^2 \le 400.\]
The \(\le\) symbol is used because this gives the set of points that are inside or on the circle.
The system will be graphed as an intersection:
\[\begin{cases}\begin{align} y &> 0 \\ x^2+y^2 &\le 400. \end{align}\end{cases}\]
\(_\square\)
