Equations and Inequalities
Key Terms
Equation - A mathematical expression that contains an equals sign.
Inequality - A mathematical expression that contains inequality signs.
Variables - Unknown values usually represented by letters.
Constants - Quantities that remain unchanged .
System of Equations - A collection of two or more equations with a same set of unknowns.
Equations and Inequalities
In algebra, an equation is a mathematical expression that contains an equals sign. It tells us that two expressions represent the same number. For example, y = 12x is an equation. An inequality is a mathematical expression that contains inequality signs. For example, y≤ 12x is an inequality. Inequalities are used to tell us that an expression is either larger or smaller than another expression. Equations and inequalities can contain both variables and constants. Variables are usually given a letter and they are used to represent unknown values. These quantities can change because they depend on other numbers in the problem. Constants are quantities that remain unchanged. Ordinary numbers like 2, -3¾, and ∏ are constants. Equations and inequalities are used as a shorthand notation for situations that involve numerical data. They are very useful because most problems require several steps to arrive at a solution, and it becomes tedious to repeatedly write out the situation in words.
Here are some examples of equations:
3x - 2 = 5
x + 9 = 2x + 5
x /3 = 15
x² + 1 = 10
To write an inequality, we use the following symbols:
Symbol | Meaning | Example | Read As |
< | Less than | x < 5 | x is less than 5 |
< | Less than or equal to | x < 5 | x is less than or equal to 5 |
> | Greater than | x > 5 | x is greater than 5 |
> | Greater than or equal to | x > 5 | x is greater than or equal to 5 |
Here are some examples of inequalities:
3x < 5
4 - x ≤ 2x
x² + 2x - 1 > 10
3x/4 ≥ x/2 - 3
The most important skill in algebra is the ability to translate a word problem into the correct equation or inequality so you can find the solution easily. The first two steps are defining the variables and translating the word problem into a mathematical equation.
Defining the variables means that we assign letters to any unknown quantities in the problem.
Translating means that we change the word expression into a mathematical expression containing variables and mathematical operations with an equal sign or an inequality sign.
Checking Solutions
For Equations
You will often need to check solutions to equations in order to check your work. In a math class, checking that you arrived at the correct solution is very good practice. We check the solution to an equation by replacing the variable in an equation with the value of the solution. A solution should result in a true statement when plugged into the equation.
For Inequalities
To check the solution to an inequality, we replace the variable in the inequality with the value of the solution. A solution to an inequality produces a true statement when substituted into the inequality.
Let's use what we have learned about defining variables, writing equations and writing inequalities to solve some real-world problems.
Example 1
Tomatoes cost $0.50 each and avocados cost $2.00 each. Anne buys six more tomatoes than avocados. Her total bill is $8. How many tomatoes and how many avocados did Anne buy?
Step 1: Define your variable
Let a = the number of avocados Anne buys.
Step 2: Translate
Anne buys six more tomatoes than avocados. This means that a + 6 = the number of tomatoes. Tomatoes cost $0.50 each and avocados cost $2.00 each. Her total bill is $8. This means that .50 times the number of tomatoes plus 2 times the number of avocados equals 8.
Remember that a = the number of avocados, so Anne buys two avocados.
The number of tomatoes is: a + 6 = 2 + 6 = 8.
Step 3: Answer
Anne bought 2 avocados and 8 tomatoes.
Step 4: Check
If Anne bought two avocados and eight tomatoes, the total cost is: (2 2) + (8 0:5) = 4 + 4 = 8. The answer checks out.
Example 2
To organize a picnic Peter needs at least two times as many hamburgers as hot dogs. He has 24 hot dogs. What is the possible number of hamburgers Peter has?
Step 1: Define your variable
Let x = number of hamburgers
Step 2: Translate
Peter needs at least two times as many hamburgers as hot dogs. He has 24 hot dogs. This means that twice the number of hot dogs is less than or equal to the number of hamburgers.
Step 3: Answer
Peter needs at least 48 hamburgers.
Step 4: Check
48 hamburgers is twice the number of hot dogs. So more than 48 hamburgers is more than twice the number of hot dogs. The answer checks out.
Linear Inequalities
Solving Inequalities
To begin, click on "presentation" below to learn about inequalities with one variable.
Dana has a budget of $350 to spend on a rental car for an upcoming trip, but she wants to spend as little of that money as possible. If the trip will last five days, what range of daily rental rates should she be willing to consider?
Like equations, inequalities show a relationship between two expressions. We solve and graph inequalities in a similar way to equations—but when we solve an inequality, the answer is usually a set of values instead of just one value.
When writing inequalities we use the following symbols:
Symbol | Meaning | Example | Read As |
< | Less than | x < 5 | x is less than 5 |
< | Less than or equal to | x < 5 | x is less than or equal to 5 |
> | Greater than | x > 5 | x is greater than 5 |
> | Greater than or equal to | x > 5 | x is greater than or equal to 5 |
Write and Graph Inequalities in One Variable on a Number Line
Let's start with the simple inequality x > 3. We read this inequality as ''x is greater than 3." The solution is the set of all real numbers that are greater than three. We often represent the solution set of an inequality with a number line graph.
Consider another simple inequality: x ≤ 4. We read this inequality as ''x is less than or equal to 4." The solution is the set of all real numbers that are equal to four or less than four. We can graph this solution set on the number line.
Notice that we use an empty circle for the endpoint of a strict inequality (like x > 3), and a filled circle for one where the equals sign is included (like x≤ 4).
Solving Inequalities
To begin watch the video below to learn how to solve inequalities using all operations:
Solving Inequalities Using Addition and Subtraction
To solve an inequality we must isolate the variable on one side of the inequality sign. To isolate the variable, we use the same basic techniques used in solving equations. We can solve some inequalities by adding or subtracting a constant from one side of the inequality.
Solving Inequalities Using Multiplication and Division
We can also solve inequalities by multiplying or dividing both sides by a constant.
For example, to solve the inequality 5x < 3, we would divide both sides by 5 to get x < 35 .
However, something different happens when we multiply or divide by a negative number. We know, for example, that 5 is greater than 3. But if we multiply both sides of the inequality 5 > 3 by -2, we get -10 > -6. And we know that's not true; -10 is less than -6.
This happens whenever we multiply or divide an inequality by a negative number, and so we have to flip the sign around to make the inequality true.
For example, to multiply 2 < 4 by -3, first we multiply the 2 and the 4 each by -3, and then we change the < sign to a > sign, so we end up with -6 > -12. The same principle applies when the inequality contains variables.
Solving Multi-Step Inequalities
In the last two sections, we considered very simple inequalities which required one step to obtain the solution. However, most inequalities require several steps to arrive at the solution. As with solving equations, we must use the order of operations to find the correct solution. In addition, remember that when we multiply or divide the inequality by a negative number, the direction of the inequality changes. The general procedure for solving multi-step inequalities is almost exactly like the procedure for solving multi-step equations:
1. Clear parentheses on both sides of the inequality and collect like terms.
2. Add or subtract terms so the variable is on one side and the constant is on the other side of the inequality sign.
3. Multiply and divide by whatever constants are attached to the variable. Remember to change the direction of the inequality if you multiply or divide by a negative number.
Example 1
Step 1: Add 3x to both sides (this is so you can combine your like terms.. you want to combine 3x with 9x)
Step 2: Simplify (you would need to put 3x/1 and then find the common denominator of 5. This would make the top number 15x and you would then add 9x/5 to 15x/5)
Step 3: Add 7 to both sides (Remember the goal is to isolate the variable!)
Step 4: Simplify
Step 5: Multiply 5 to both sides
Step 6: Simplify
Step 7: Divide both sides by 24
Step 8: Simplify to get your final answer
Step 9: Graph it on a number line
Example 2
Step 1: Add 10x to both sides
Step 2: Simplify
Step 3: Subtract 12 from both sides
Step 4: Simplify
Step 5: Divide both sides by -15 (NOTICE: You are dividing by a negative number so we FLIP the sign!)
Step 6: Simplify for your final answer
Step 7: Graph it on a number line (NOTICE: We graph between 1 and 2 because 8/5 reduced is 1 and 3/5ths.)
Using Inequalities
Expressing Solutions of an Inequality
The solution of an inequality can be expressed in four different ways:
Inequality Notation
The answer is simply expressed as x < 15.
Set Notation
The answer is expressed as a set: {x|x < 15g}. The brackets indicate a set and the vertical line means ''such that," so we read this expression as ''the set of all values of x such that x is a real number less than 15".
Interval Notation
Uses brackets to indicate the range of values in the solution. For example, the answer to our problem would be expressed as (-∞,15), meaning ''the interval containing all the numbers from -∞ to 15 but not actually including - ∞ or 15".
(a) Square or closed brackets ''[" and '']" indicate that the number next to the bracket is included in the solution set.
(b) Round or open brackets ''(" and '')" indicate that the number next to the bracket is not included in the solution set. When using infinity and negative infinity (∞ and -∞), we always use open brackets, because infinity isn't an actual number and so it can't ever really be included in an interval.
Solution Graph
Shows the solution on the real number line. A closed circle on a number indicates that the number is included in the solution set, while an open circle indicates that the number is not included in the set. For our example, the solution graph is:
Identify the Number of Solutions of an Inequality
Inequalities can have:
A set that has an infinite number of solutions.
A set that has a discrete number of solutions.
No solutions.
Infinite Number of Solutions
Discrete Number of Solutions
However, in real life, sometimes we are trying to solve a problem that can only have positive integer answers, because the answers describe numbers of discrete objects.
For example, suppose you are trying to figure out how many $8 CDs you can buy if you want to spend less than $50. An inequality to describe this situation would be 8x < 50, and if you solved that inequality you would get x < 50/8 or x < 6.25.
But could you really buy any number of CDs as long as its less than 6.25? No; you couldn't really buy 6.1 CDs, or -5 CDs, or any other fractional or negative number of CDs. So if we wanted to express our solution in set notation, we couldn't express it as the set of all numbers less than 6.25, or {x|x < 6:25}.
Instead, the solution is just the set containing all the nonnegative whole numbers less than 6.25, or {0, 1, 2, 3, 4, 5, 6}. When were solving a real-world problem dealing with discrete objects like CDs, our solution set will often be a finite set of numbers instead of an infinite interval.
No Solutions
An inequality can also have no solutions at all. For example, consider the inequality x -5 > x + 6. When we subtract x from both sides, we end up with -5 > 6, which is not true for any value of x. We say that this inequality has no solution.
The opposite can also be true. If we flip the inequality sign in the above inequality, we get x - 5, which simplifies to -5 < 6. That's always true no matter what x is, so the solution to that inequality would be all real numbers, or (-∞,∞). Solve Real-World Problems Using Inequalities
Solving real-world problems that involve inequalities is very much like solving problems that involve equations.
Solve Real-World Problems Using Inequalities
Solving real-world problems that involve inequalities is very much like solving problems that involve equations.
Example 1
In order to get a bonus this month, Leon must sell at least 120 newspaper subscriptions. He sold 85 subscriptions in the first three weeks of the month. How many subscriptions must Leon sell in the last week of the month?
Solution
Let x = the number of subscriptions Leon sells in the last week of the month. The total number of subscriptions for the month must be greater than 120, so we write :
85 + x ≥ 120.
We solve the inequality by subtracting 85 from both sides: x ≥ 35.
Leon must sell 35 or more subscriptions in the last week to get his bonus.
Check
To check the answer, we see that 85 + 35 = 120. If he sells 35 or more subscriptions, the total number of subscriptions he sells that month will be 120 or more. The answer checks out.
Example 2
Virenas Scout troop is trying to raise at least $650 this spring. How many boxes of cookies must they sell at $4.50 per box in order to reach their goal?
Solution
Let x = number of boxes sold. Then the inequality describing this problem is 4.50 ≥ 650.
We solve the inequality by dividing both sides by 4.50: x ≥ 144.44.
We round up the answer to 145 since only whole boxes can be sold.
Virenas troop must sell at least 145 boxes.
Check
If we multiply 145 by $4.50 we obtain $652.50, so if Virenas troop sells more than 145 boxes they will raise more than $650. But if they sell 144 boxes, they will only raise $648,
which is not enough. So they must indeed sell at least 145 boxes. The answer checks out.
Example 3
The width of a rectangle is 20 inches. What must the length be if the perimeter is at least 180 inches?
Solution
Let x = length of the rectangle. The formula for perimeter is:
Perimeter = 2 x length + 2 x width
Since the perimeter must be at least 180 inches, we have 2x + 2(20) ≥ 180.
Simplify: 2x + 40 ≥ 180
Subtract 40 from both sides: 2x ≥ 140
Divide both sides by 2: x ≥ 70
The length must be at least 70 inches.
Check
If the length is at least 70 inches and the width is 20 inches, then the perimeter is at least 2(70)+2(20) =180 inches. The answer checks out.
Linear Inequalities in Two Variables
Yasmeen is selling handmade bracelets for $5 each and necklaces for $7 each. How many of both does she need to sell to make at least $100?
A linear inequality in two variables takes the form y > mx+b or y. Linear inequalities are closely related to graphs of straight lines; recall that a straight line has the equation y =mx + b.
When we graph a line in the coordinate plane, we can see that it divides the plane in half:
The solution to a linear inequality includes all the points in one half of the plane. We can tell which half by looking at the inequality sign:
Inequality Sign | Solution Set |
> | The solution set is the half plane above the line. |
≥ | The solution set is the half plane above the line and also all the points on the line. |
< | The solution set is the half plane below the line. |
≤ | The solution set is the half plane below the line and also all the points on the line. |
For a strict inequality, we draw a dashed line to show that the points in the line are not part of the solution. For an inequality that includes the equals sign, we draw a solid line to show that the points on the line are part of the solution.
Here are some examples of linear inequality graphs.
Graph Linear Inequalities in One Variable in the Coordinate Plane
In the last few sections we graphed inequalities in one variable on the number line. We can also graph inequalities in one variable on the coordinate plane. We just need to remember that when we graph an equation of the type x = a we get a vertical line, and when we graph an equation of the type y = b we get a horizontal line.
Example 1
Graph the inequality x > 4 on the coordinate plane.
Solution
First let's remember what the solution to x > 4 looks like on the number line.
The solution to this inequality is the set of all real numbers x that are bigger than 4, not including 4. The solution is represented by a line.
In two dimensions, the solution still consists of all the points to the right of x = 4, but for all possible y-values as well. This solution is represented by the half plane to the right of x = 4. (You can think of it as being like the solution graphed on the number line, only stretched out vertically.)
The line x = 4 is dashed because the equals sign is not included in the inequality, meaning that points on the line are not included in the solution.
Example 2
Graph the inequality |x| ≥ 2.
Solution
The absolute value inequality |x| ≥ 2 can be re-written as a compound inequality: x ≤ -2 or x ≥ 2
In other words, the solution is all the coordinate points for which the value of x is smaller than or equal to -2 or greater than or equal to 2. The solution is represented by the plane to the left of the vertical line x = -2 and the plane to the right of line x = 2.
Both vertical lines are solid because points on the lines are included in the solution.
Example 3
Graph the inequality |y| < 5
Solution
The absolute value inequality |y| < 5 can be re-written as -5. This is a compound inequality which can be expressed as: y > -5 and y < 5
In other words, the solution is all the coordinate points for which the value of y is larger than -5 and smaller than 5. The solution is represented by the plane between the horizontal lines y = -5 and y = 5.
Both horizontal lines are dashed because points on the lines are not included in the solution.
Graph Linear Inequalities in Two Variables
The general procedure for graphing inequalities in two variables is as follows:
1. Re-write the inequality in slope-intercept form: y = mx + b. Writing the inequality in this form lets you know the direction of the inequality.
2. Graph the line of the equation y = mx + b using your favorite method (plotting two points, using slope and y-intercept, using y-intercept and another point, or whatever is easiest)
Draw the line as a dashed line if the equals sign is not included and a solid line if the equals sign is included.
3. Shade the half plane above the line if the inequality is ''greater than". Shade the half plane under the line if the inequality is ''less than".
Watch the "presentation" below to learn more.
Example 1
Graph the inequality y ≥ 2x - 3.
Solution
The inequality is already written in slope-intercept form, so its easy to graph. First we graph the line y = 2x - 3; then we shade the half-plane above the line. The line is solid because the inequality includes the equals sign.
Example 2
Graph the inequality 5x - 2y > 4.
Solution
First we need to rewrite the inequality in slope-intercept form:
-2y > -5x + 4
Notice that the inequality sign changed direction because we divided by a negative number.
To graph the equation, we can make a table of values:
x | y |
-2 | 5/2 (-2) - 2 = -7 |
0 | 5/2 (0) - 2 = -2 |
2 | 5/2 (2) - 2 = 3 |
After graphing the line, we shade the plane below the line because the inequality in slope-intercept form is less than. The line is dashed because the inequality does not include an equals sign.
Solve Real-World Problems Using Linear Inequalities
In this section, we see how linear inequalities can be used to solve real-world applications.
Example 1
A retailer sells two types of coffee beans. One type costs $9 per pound and the other type costs $7 per pound. Find all the possible amounts of the two different coffee beans that can
be mixed together to get a quantity of coffee beans costing $8.50 or less.
Solution
Let x = weight of $9 per pound coffee beans in pounds.
Let y = weight of $7 per pound coffee beans in pounds.
The cost of a pound of coffee blend is given by 9x + 7y.
We are looking for the mixtures that cost $8.50 or less. We write the inequality 9x + 7y ≤ 8.50.
Since this inequality is in standard form, its easiest to graph it by finding the x and y- intercepts.
When x = 0, we have 7y = 8.50 or y = 8.50/7≈ 1.21
When y = 0, we have 9x = 8.50 or x = 8.50/9 ≈ 0.94
We can then graph the line that includes those two points.
Now we have to figure out which side of the line to shade. In y-intercept form, we shade the area below the line when the inequality is ''less than. But in standard form that is not always true. We could convert the inequality to y-intercept form to find out which side to shade, but there is another way that can be easier.
The other method, which works for any linear inequality in any form, is to plug a random point into the inequality and see if it makes the inequality true. Any point that's not on the line will do; the point (0, 0) is usually the most convenient.
In this case, plugging in 0 for x and y would give us 9(0) + 7(0) ≤ 8.50, which is true. That means we should shade the half of the plane that includes (0, 0). If plugging in (0, 0)
gave us a false inequality, that would mean that the solution set is the part of the plane that does not contain (0, 0).
Notice also that in this graph we show only the first quadrant of the coordinate plane. That's because weight values in the real world are always nonnegative, so points outside the first quadrant dont represent real-world solutions to this problem.
Example 2
Julius has a job as an appliance salesman. He earns a commission of $60 for each washing machine he sells and $130 for each refrigerator he sells. How many washing machines and
refrigerators must Julius sell in order to make $1000 or more in commissions?
Solution
Let x = number of washing machines Julius sells.
Let y = number of refrigerators Julius sells.
The total commission is 60x + 130y.
Were looking for a total commission of $1000 or more, so we write the inequality 60x + 130y ≥ 1000.
Once again, we can do this most easily by finding the x and y-intercepts.
When x = 0, we have 130y = 1000, or y = 1000/30 ≈7.69.
When y = 0, we have 60x = 1000, or x = 1000/60≈16.67.
We draw a solid line connecting those points, and shade above the line because the inequality is ''greater than." We can check this by plugging in the point (0, 0): selling 0 washing
machines and 0 refrigerators would give Julius a commission of $0, which is not greater than or equal to $1000, so the point (0, 0) is not part of the solution; instead, we want to shade the side of the line that does not include it.
Solving Systems of Equations and Inequalities
In this lesson, we'll discover methods to determine if an ordered pair is a solution to a system of two equations. Then we'll learn to solve the two equations graphically and confirm that the solution is the point where the two lines intersect. Finally, we'll look at real-world problems that can be solved using the methods described in this chapter.
To begin, click on the "presentation" below to learn about solving systems of Equations and Inequalities.
Determine Whether an Ordered Pair is a Solution to a System of Equations
A linear system of equations is a set of equations that must be solved together to find the one solution that fits them both.
Consider this system of equations:
y = x + 2
y = -2x + 1
Since the two lines are in a system, we deal with them together by graphing them on the same coordinate axes. We can use any method to graph them; let's do it by making a table of values for each line.
Line 1: y = x + 2
x | y |
0 | 2 |
1 | 3 |
Line 2: -2x + 1
x | y |
0 | 1 |
1 | -1 |
We already know that any point that lies on a line is a solution to the equation for that line. That means that any point that lies on both lines in a system is a solution to both equations. So in this system:
Point A is not a solution to the system because it does not lie on either of the lines.
Point B is not a solution to the system because it lies only on the blue line but not on the red line.
Point C is a solution to the system because it lies on both lines at the same time.
In fact, point C is the only solution to the system, because it is the only point that lies on both lines. For a system of equations, the geometrical solution is the intersection of the two lines in the system. The algebraic solution is the ordered pair that solves both equations—in other words, the coordinates of that intersection point.
You can confirm the solution by plugging it into the system of equations, and checking that the solution works in each equation.
Example 1
Determine which of the points (1, 3), (0, 2), or (2, 7) is a solution to the following system of equations:
y = 4x - 1
y = 2x + 3
Solution
To check if a coordinate point is a solution to the system of equations, we plug each of the x and y values into the equations to see if they work.
Point (1, 3):
Point (1, 3) is on the line y = 4x -1 but it is not on the line y = 2x + 3, so it is not a solution to the system.
Point (0, 2):
Point (0, 2) is not on the line y = 4x - 1, so it is not a solution to the system. Note that it is not necessary to check the second equation because the point needs to be on both lines for it to be a solution to the system.
Point (2 , 7):
Point (2, 7) is a solution to the system since it lies on both lines. The solution to the system is the point (2, 7).
Determine the Solution to a Linear System by Graphing
The solution to a linear system of equations is the point, (if there is one) that lies on both lines. In other words, the solution is the point where the two lines intersect.
We can solve a system of equations by graphing the lines on the same coordinate plane and reading the intersection point from the graph.
This method most often offers only approximate solutions, so its not sufficient when you need an exact answer. However, graphing the system of equations can be a good way to get a sense of what's really going on in the problem you're trying to solve, especially when it's a real-world problem.
Example 1
Solve the following system of equations by graphing:
y = 3x - 5
y = -2x + 5
Solution
Graph both lines on the same coordinate axis using any method you like.
In this case, lets make a table of values for each line.
Line 1: y = 3x -5
x | y |
1 | -2 |
2 | 1 |
Line 2: y = -2x + 5
x | y |
1 | 3 |
2 | 1 |
The solution to the system is given by the intersection point of the two lines. The graph shows that the lines intersect at point (2, 1). So the solution is x = 2; y = 1 or (2, 1).
Example 2
Solve the following system of equations by graphing:
2x + 3y = 6
4x - y = -2
Solution
Since the equations are in standard form, this time we'll graph them by finding the x and y-intercepts of each of the lines.
Line 1: 2x + 3y = 6
Line 2: -4x + y = 2
The graph shows that the lines intersect at (0, 2). Therefore, the solution to the system of equations is x = 0; y = 2.
Solve Real-World Problems Using Graphs of Linear Systems
Consider the following problem:
Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?
Let's start by drawing a sketch. Here's what the race looks like when Nadia starts running; we'll call this time t = 0.
Now let's define two variables in this problem:
t = the time from when Nadia starts running
d = the distance of the runners from the starting point.
Since there are two runners, we need to write equations for each of them. That will be the system of equations for this problem.
For each equation, we use the formula: distance = speed × time
Nadia's equation: d = 6t
Peter's equation: d = 5t + 20
(Remember that Peter was already 20 feet from the starting point when Nadia started running.)
Let's graph these two equations on the same coordinate axes.
Time should be on the horizontal axis since it is the independent variable. Distance should be on the vertical axis since it is the dependent variable.
We can use any method for graphing the lines, but in this case we'll use the slope–intercept method since it makes more sense physically.
To graph the line that describes Nadia's run, start by graphing the y-intercept: (0, 0). (If you don't see that this is the y-intercept, try plugging in the test-value of x = 0.)
The slope tells us that Nadia runs 6 feet every one second, so another point on the line is (1, 6). Connecting these points gives us Nadia's line:
To graph the line that describes Peter's run, again start with the y-intercept. In this case this is the point (0, 20).
The slope tells us that Peter runs 5 feet every one second, so another point on the line is (1, 25). Connecting these points gives us Peter's line:
In order to find when and where Nadia and Peter meet, we'll graph both lines on the same graph and extend the lines until they cross. The crossing point is the solution to this problem.
The graph shows that Nadia and Peter meet 20 seconds after Nadia starts running, and 120 feet from the starting point.
Solving Linear Systems by Substitution
To begin, watch the "presentation" below to learn about solving linear systems by substitution.
Let's look again at the problem about Peter and Nadia racing.
Peter and Nadia like to race each other. Peter can run at a speed of 5 feet per second and Nadia can run at a speed of 6 feet per second. To be a good sport, Nadia likes to give Peter a head start of 20 feet. How long does Nadia take to catch up with Peter? At what distance from the start does Nadia catch up with Peter?
In that example we came up with two equations:
Nadia's equation: d = 6t
Peter's equation: d = 5t + 20
Each equation produced its own line on a graph, and to solve the system we found the point at which the lines intersected—the point where the values for d and t satisfied both relationships. When the values for d and t are equal, that means that Peter and Nadia are at the same place at the same time. But there's a faster way than graphing to solve this system of equations. Since we want the value of d to be the same in both equations, we could just set the two right-hand sides of the equations equal to each other to solve for t. That is, if d = 6t and d = 5t+20, and the two d's are equal to each other, then by the transitive property we have 6t = 5t + 20. We can solve this for t:
Even if the equations weren't so obvious, we could use simple algebraic manipulation to find an expression for one variable in terms of the other. If we rearrange Peter's equation to isolate t:
We can now substitute this expression for t into Nadia's equation (d = 6t) to solve:
So we find that Nadia and Peter meet 20 seconds after they start racing, at a distance of 120 feet away. The method we just used is called the Substitution Method. In this lesson you'll learn several techniques for isolating variables in a system of equations, and for using those expressions to solve systems of equations that describe situations like this one.
Example 1
Let's look at an example where the equations are written in standard form.
Solve the system:
2x + 3y = 6
-4x + y = 2
Again, we start by looking to isolate one variable in either equation. If you look at the second equation, you should see that the coefficient of y is 1. So the easiest way to start is to use this equation to solve for y.
As you can see, we end up with the same solution (x = 0; y = 2) that we found when we graphed these functions in a previous lesson. So, as long as you are careful with the algebra, the substitution method can be a very efficient way to solve systems. Next, let's look at a more complicated example. Here, the values of x and y we end up with aren't whole numbers so they would be more difficult to graph.
Solving Real-World Problems Using Linear Systems
Simultaneous equations can help us solve many real-world problems. We may be considering a purchase—for example, trying to decide whether it's cheaper to buy an item online where you pay shipping or at the store where you do not. Or you may wish to join a CD music club, but aren't sure if you would really save any money by buying a new CD every month in that way. Or you might be considering two different phone contracts. Let's look at an example of that now.
Example 1
Anne is trying to choose between two phone plans. The first plan, with Vendafone, costs $20 per month, with calls costing an additional 25 cents per minute. The second company, Sellnet, charges $40 per month, but calls cost only 8 cents per minute. Which should she choose?
You should see that Anne's choice will depend upon how many minutes of calls she expects to use each month. We start by writing two equations for the cost in dollars in terms of the minutes used. Since the number of minutes is the independent variable, it will be our x.
Cost is dependent on minutes - the cost per month is the dependent variable and will be assigned y.
For Vendafone: y = 0.25x + 20
For Sellnet: y = 0.08x + 40
By writing the equations in slope-intercept form (y = mx + b), you can sketch a graph to visualize the
situation:
The line for Vendafone has an intercept of 20 and a slope of 0.25. The Sellnet line has an intercept of 40 and a slope of 0.08 (which is roughly a third of the Vendafone line's slope). In order to help Anne decide which to choose, we'll find where the two lines cross, by solving the two equations as a system.
Since equation 1 gives us an expression for y(0.25x + 20), we can substitute this expression directly into equation 2:
So if Anne uses 117.65 minutes a month (although she can't really do exactly that, because phone plans only count whole numbers of minutes), the phone plans will cost the same. Now we need to look at the graph to see which plan is better if she uses more minutes than that, and which plan is better if she uses
fewer. You can see that the Vendafone plan costs more when she uses more minutes, and the Sellnet plan costs more with fewer minutes.
So, if Anne will use 117 minutes or less every month she should choose Vendafone. If she plans on using 118 or more minutes she should choose Sellnet.
Mixture Problems
To begin, watch the "presentation" below:
Systems of equations crop up frequently in problems that deal with mixtures of two things—chemicals in a solution, nuts and raisins, or even the change in your pocket! Let's look at some examples of these.
Example 1
Janine empties her purse and finds that it contains only nickels (worth 5 cents each) and dimes (worth 10 cents each). If she has a total of 7 coins and they have a combined value of 45 cents, how many of each coin does she have?
Since we have 2 types of coins, let's call the number of nickels x and the number of dimes y. We are given two key pieces of information to make our equations: the number of coins and their value.
Number of Coins in Equation | x + y = 7 | (number of nickels) + (number of dimes) |
Value Equation | 5x + 10y = 55 | Since nickels are worth 5 cents and dimes are worth 10 cents. |
We can quickly rearrange the first equation to isolate x:
Janine has 3 nickels and 4 dimes.
Solving Linear Systems by Elimination
To begin, watch the "presentation" below to learn about solving linear systems by elimination.
In this lesson, we'll see how to use simple addition and subtraction to simplify our system of equations to a single equation involving a single variable. Because we go from two unknowns (x and y) to a single unknown (either x or y), this method is often referred to by solving by elimination. We eliminate one variable in order to make our equations solvable! To illustrate this idea, let's look at the simple example of buying apples and bananas.
Example 1
If one apple plus one banana costs $1.25 and one apple plus 2 bananas costs $2.00, how much does one banana cost? One apple?
It shouldn't take too long to discover that each banana costs $0.75. After all, the second purchase just contains 1 more banana than the first, and costs $0.75 more, so that one banana must cost $0.75.
Here's what we get when we describe this situation with algebra:
a + b = 1.25
a + 2b = 2.00
Now we can subtract the number of apples and bananas in the first equation from the number in the second equation, and also subtract the cost in the first equation from the cost in the second equation, to get the difference in cost that corresponds to the difference in items purchased.
(a + 2b) - (a + b) = 2.00 - 1.25
so b = 0.75
That gives us the cost of one banana. To find out how much one apple costs, we subtract $0.75 from the total cost of one apple and one banana.
a + 0.75 = 1:25
a = 1.25 - 0.75
a = 0.50
So an apple costs 50 cents. To solve systems using addition and subtraction, we'll be using exactly this idea – by looking at the sum or difference of the two equations we can determine a value for one of the unknowns.
Solving Linear Systems Using Addition of Equations
Often considered the easiest and most powerful method of solving systems of equations, the addition (or elimination) method lets us combine two equations in such a way that the resulting equation has only one variable. We can then use simple algebra to solve for that variable. Then, if we need to, we can substitute the value we get for that variable back into either one of the original equations to solve for the other variable.
Example 1
Solve this system by addition:
3x + 2y = 11
5x - 2y = 13
Solution
We will add everything on the left of the equals sign from both equations, and this will be equal to the sum of everything on the right:
(3x + 2y) + (5x - 2y) = 11 + 13
8x = 24
x = 3
A simpler way to visualize this is to keep the equations as they appear above, and to add them together vertically, going down the columns. However, just like when you add units, tens and hundreds, you MUST be sure to keep the x's and y's in their own columns. You may also wish to use terms like "0y" as a placeholder!
Again we get 8x = 24, or x = 3. To find a value for y, we simply substitute our value for x back in.
Substitute x = 3 into the second equation:
The reason this method worked is that the y-coefficients of the two equations were opposites of each other: 2 and -2.
Because they were opposites, they canceled each other out when we added the two equations together, so our final equation had no y-term in it and we could just solve it for x.
In a little while we'll see how to use the addition method when the coefficients are not opposites, but for now let's look at another example where they are.
Comparing Methods for Solving Linear Systems
Now that we've covered the major methods for solving linear equations, let's review them. For simplicity, we'll look at them in table form. This should help you decide which method would be best for a given situation.
Method | Best Used When You... | Advantages | Comments |
Graphing | .. don't need an accurate answer. | Often easier to see number and quality of intersections on a graph. With a graphing calculator, it can be the fastest method since you don't have to do any computations. | Can lead to imprecise answers with non-integer solutions. |
Substitution | ... have an explicit equation for one variable (e.g. y = 14x + 2) | Works on all systems. Reduces the system to one variable, making it easier to solve. | You are not often given explicit functions in systems problems, so you may have to do extra work to get one o f the equations into that form. |
Elimination by Addition or Subtraction | ...have matching coefficients for one variable in both equations. | Easy to combine equations to eliminate one variable. Quick to solve. | It is not very likely that a given system will have matching coefficients. |
Elimination by Multiplication and then Addition and Subtraction | ...do not have any variables defined explicitly or any matching coefficients. | Works on all systems. Makes it possible to combine equations to eliminate one variable. | Often more algebraic manipulation is needed to prepare the equations. |
Special Types of Linear Systems
A system of linear equations is a set of linear equations which must be solved together. The lines in the system can be graphed together on the same coordinate graph and the solution to the system is the point at which the two lines intersect.
Or at least that's what usually happens. But what if the lines turn out to be parallel when we graph them?
If the lines are parallel, they won't ever intersect. That means that the system of equations they represent has no solution. A system with no solutions is called an inconsistent system.
And what if the lines turn out to be identical?
If the two lines are the same, then every point on one line is also on the other line, so every point on the line is a solution to the system. The system has an infinite number of solutions, and the two equations are really just different forms of the same equation. Such a system is called a dependent system .
But usually, two lines cross at exactly one point and the system has exactly one solution:
A system with exactly one solution is called a consistent system.
To identify a system as consistent, inconsistent, or dependent, we can graph the two lines on the same graph and see if they intersect, are parallel, or are the same line. But sometimes
it is hard to tell whether two lines are parallel just by looking at a roughly sketched graph.
Another option is to write each line in slope-intercept form and compare the slopes and y- intercepts of the two lines. To do this we must remember that:
Lines with different slopes always intersect.
Lines with the same slope but different y-intercepts are parallel.
Lines with the same slope and the same y-intercepts are identical.
We can also identify a system algebraically:
A consistent system will always give exactly one solution.
An inconsistent system will yield a statement that is always false (like 0 = 13).
A dependent system will yield a statement that is always true (like 9 = 9).
Systems of Linear Inequalities
To begin, watch the "presentation" below to learn about systems of linear inequalities.
In the last section, you learned how to graph a linear inequality in two variables. To do that, you graphed the equation of the straight line on the coordinate plane. The line was solid for ≤ or ≥ signs (where the equals sign is included), and the line was dashed for < or > signs (where the equals sign is not included).
Then you shaded above the line (if the inequality began with y > or y ≥ ) or below the line (if it began with y < or y ≤).
In this section, we'll see how to graph two or more linear inequalities on the same coordinate plane. The inequalities are graphed separately on the same graph, and the solution for the system is the common shaded region between all the inequalities in the system. One linear inequality in two variables divides the plane into two half-planes. A system of two or more linear inequalities can divide the plane into more complex shapes.
Lets start by solving a system of two inequalities.
Graph a System of Two Linear Inequalities
Example 1
Solve the following system:
2x + 3y ≤ 18
x - 4y ≤ 12
Solution
Solving systems of linear inequalities means graphing and finding the intersections. So we graph each inequality, and then find the intersection regions of the solution.
First, let's rewrite each equation in slope-intercept form. (Remember that this form makes it easier to tell which region of the coordinate plane to shade.) Our system becomes
Notice that the inequality sign in the second equation changed because we divided by a negative number! For this first example, we'll graph each inequality separately and then combine the results. Here's the graph of the first inequality:
The line is solid because the equals sign is included in the inequality. Since the inequality is less than or equal to, we shade below the line.
And here's the graph of the second inequality:
The line is solid again because the equals sign is included in the inequality. We now shade above the line because y is greater than or equal to.
When we combine the graphs, we see that the blue and red shaded regions overlap. The area where they overlap is the area where both inequalities are true. Thus that area (shown below in purple) is the solution of the system.
The kind of solution displayed in this example is called unbounded, because it continues forever in at least one direction (in this case, forever upward and to the left).
Example 2
There are also situations where a system of inequalities has no solution. For example, let's solve this system.
y ≤ 2x -4
y > 2x + 6
Solution
We start by graphing the first line. The line will be solid because the equals sign is included in the inequality. We must shade downwards because y is less than.
Next we graph the second line on the same coordinate axis. This line will be dashed because the equals sign is not included in the inequality. We must shade upward because y is greater than.
It doesn't look like the two shaded regions overlap at all. The two lines have the same slope, so we know they are parallel; that means that the regions indeed won't ever overlap since the lines won't ever cross.
So this system of inequalities has no solution.
But a system of inequalities can sometimes have a solution even if the lines are parallel. For example, what happens if we swap the directions of the inequality signs in the system we just graphed?
To graph the system
y ≥ 2x -4
y < 2x + 6
we draw the same lines we drew for the previous system, but we shade upward for the first inequality and downward for the second inequality. Here is the result
You can see that this time the shaded regions overlap. The area between the two lines is the solution to the system.
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