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# Systems of Linear Equations: Two Variables

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible? In this section, we will consider linear equations with two variables to answer these and similar questions.
(credit: Thomas Sørenes)

## Introduction to Solutions of Systems

In order to investigate situations such as that of the skateboard manufacturer, we need to recognize that we are dealing with more than one variable and likely more than one equation. A system of linear equations consists of two or more linear equations made up of two or more variables such that all equations in the system are considered simultaneously. To find the unique solution to a system of linear equations, we must find a numerical value for each variable in the system that will satisfy all equations in the system at the same time. Some linear systems may not have a solution and others may have an infinite number of solutions. In order for a linear system to have a unique solution, there must be at least as many equations as there are variables. Even so, this does not guarantee a unique solution. In this section, we will look at systems of linear equations in two variables, which consist of two equations that contain two different variables. For example, consider the following system of linear equations in two variables.

$\begin{array}{c}2x+y=\text{ }15\\ 3x-y=\text{ }5\end{array}$

The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently. In this example, the ordered pair (4, 7) is the solution to the system of linear equations. We can verify the solution by substituting the values into each equation to see if the ordered pair satisfies both equations. Shortly we will investigate methods of finding such a solution if it exists.

$\begin{array}{l}2\left(4\right)+\left(7\right)=15\text{ }\text{True}\hfill \\ 3\left(4\right)-\left(7\right)=5\text{ }\text{True}\hfill \end{array}$

In addition to considering the number of equations and variables, we can categorize systems of linear equations by the number of solutions. A consistent system of equations has at least one solution. A consistent system is considered to be an independent system if it has a single solution, such as the example we just explored. The two lines have different slopes and intersect at one point in the plane. A consistent system is considered to be a dependent system if the equations have the same slope and the same y-intercepts. In other words, the lines coincide so the equations represent the same line. Every point on the line represents a coordinate pair that satisfies the system. Thus, there are an infinite number of solutions. Another type of system of linear equations is an inconsistent system, which is one in which the equations represent two parallel lines. The lines have the same slope and different y-intercepts. There are no points common to both lines; hence, there is no solution to the system.

### A General Note: Types of Linear Systems

There are three types of systems of linear equations in two variables, and three types of solutions.
• An independent system has exactly one solution pair $\left(x,y\right)$. The point where the two lines intersect is the only solution.
• An inconsistent system has no solution. Notice that the two lines are parallel and will never intersect.
• A dependent system has infinitely many solutions. The lines are coincident. They are the same line, so every coordinate pair on the line is a solution to both equations.
Below is a comparison of graphical representations of each type of system.

### How To: Given a system of linear equations and an ordered pair, determine whether the ordered pair is a solution.

1. Substitute the ordered pair into each equation in the system.
2. Determine whether true statements result from the substitution in both equations; if so, the ordered pair is a solution.

### Example: Determining Whether an Ordered Pair Is a Solution to a System of Equations

Determine whether the ordered pair $\left(5,1\right)$ is a solution to the given system of equations.

$\begin{array}{l}x+3y=8\hfill \\ 2x - 9=y\hfill \end{array}$

Answer: Substitute the ordered pair $\left(5,1\right)$ into both equations.

$\begin{array}{ll}\left(5\right)+3\left(1\right)=8\hfill & \hfill \\ \text{ }8=8\hfill & \text{True}\hfill \\ 2\left(5\right)-9=\left(1\right)\hfill & \hfill \\ \text{ }\text{1=1}\hfill & \text{True}\hfill \end{array}$ The ordered pair $\left(5,1\right)$ satisfies both equations, so it is the solution to the system.

#### Analysis of the Solution

We can see the solution clearly by plotting the graph of each equation. Since the solution is an ordered pair that satisfies both equations, it is a point on both of the lines and thus the point of intersection of the two lines.

### Try It

Determine whether the ordered pair $\left(8,5\right)$ is a solution to the following system.

$\begin{array}{c}5x - 4y=20\\ 2x+1=3y\end{array}$

### Solving Systems of Equations by Graphing

There are multiple methods of solving systems of linear equations. For a system of linear equations in two variables, we can determine both the type of system and the solution by graphing the system of equations on the same set of axes.

### Example: Solving a System of Equations in Two Variables by Graphing

Solve the following system of equations by graphing. Identify the type of system.

$\begin{array}{c}2x+y=-8\\ x-y=-1\end{array}$

Answer: Solve the first equation for $y$.

$\begin{array}{c}2x+y=-8\\ y=-2x - 8\end{array}$

Solve the second equation for $y$.

$\begin{array}{c}x-y=-1\\ y=x+1\end{array}$

Graph both equations on the same set of axes: The lines appear to intersect at the point $\left(-3,-2\right)$. We can check to make sure that this is the solution to the system by substituting the ordered pair into both equations.

$\begin{array}{ll}2\left(-3\right)+\left(-2\right)=-8\hfill & \hfill \\ \text{ }-8=-8\hfill & \text{True}\hfill \\ \text{ }\left(-3\right)-\left(-2\right)=-1\hfill & \hfill \\ \text{ }-1=-1\hfill & \text{True}\hfill \end{array}$

The solution to the system is the ordered pair $\left(-3,-2\right)$, so the system is independent.

### Try It

Solve the following system of equations by graphing.

$\begin{array}{l}\text{ }2x - 5y=-25\hfill \\ -4x+5y=35\hfill \end{array}$

Answer: The solution to the system is the ordered pair $\left(-5,3\right)$.

### Q& A

#### Can graphing be used if the system is inconsistent or dependent?

Yes, in both cases we can still graph the system to determine the type of system and solution. If the two lines are parallel, the system has no solution and is inconsistent. If the two lines are identical, the system has infinite solutions and is a dependent system.

## The Substitution and Addition Methods

### Solving Systems of Equations by Substitution

Solving a linear system in two variables by graphing works well when the solution consists of integer values, but if our solution contains decimals or fractions, it is not the most precise method. We will consider two more methods of solving a system of linear equations that are more precise than graphing. One such method is solving a system of equations by the substitution method, in which we solve one of the equations for one variable and then substitute the result into the second equation to solve for the second variable. Recall that we can solve for only one variable at a time, which is the reason the substitution method is both valuable and practical.

### How To: Given a system of two equations in two variables, solve using the substitution method.

1. Solve one of the two equations for one of the variables in terms of the other.
2. Substitute the expression for this variable into the second equation, then solve for the remaining variable.
3. Substitute that solution into either of the original equations to find the value of the first variable. If possible, write the solution as an ordered pair.
4. Check the solution in both equations.

### Example: Solving a System of Equations in Two Variables by Substitution

Solve the following system of equations by substitution. [latex-display]\begin{array}{l}\text{ }-x+y=-5\hfill \\ \text{ }2x - 5y=1\hfill \end{array}[/latex-display]

Answer: First, we will solve the first equation for $y$. [latex-display]\begin{array}{l}-x+y=-5\hfill \\ \text{ }y=x - 5\hfill \end{array}[/latex-display] Now we can substitute the expression $x - 5$ for $y$ in the second equation. [latex-display]\begin{array}{l}\text{ }2x - 5y=1\hfill \\ 2x - 5\left(x - 5\right)=1\hfill \\ \text{ }2x - 5x+25=1\hfill \\ \text{ }-3x=-24\hfill \\ \text{ }x=8\hfill \end{array}[/latex-display] Now, we substitute $x=8$ into the first equation and solve for $y$. [latex-display]\begin{array}{l}-\left(8\right)+y=-5\hfill \\ \text{ }y=3\hfill \end{array}[/latex-display] Our solution is $\left(8,3\right)$. Check the solution by substituting $\left(8,3\right)$ into both equations. [latex-display]\begin{array}{llll}-x+y=-5\hfill & \hfill & \hfill & \hfill \\ -\left(8\right)+\left(3\right)=-5\hfill & \hfill & \hfill & \text{True}\hfill \\ 2x - 5y=1\hfill & \hfill & \hfill & \hfill \\ 2\left(8\right)-5\left(3\right)=1\hfill & \hfill & \hfill & \text{True}\hfill \end{array}[/latex-display]

### Try It

Solve the following system of equations by substitution. [latex-display]\begin{array}{l}x=y+3\hfill \\ 4=3x - 2y\hfill \end{array}[/latex-display]

Answer: $\left(-2,-5\right)$

### Example: Using the Addition Method When Multiplication of Both Equations Is Required

Solve the given system of equations in two variables by addition. [latex-display]\begin{array}{c}2x+3y=-16\\ 5x - 10y=30\end{array}[/latex-display]

Answer: One equation has $2x$ and the other has $5x$. The least common multiple is $10x$ so we will have to multiply both equations by a constant in order to eliminate one variable. Let’s eliminate $x$ by multiplying the first equation by $-5$ and the second equation by $2$. [latex-display]\begin{array}{l} -5\left(2x+3y\right)=-5\left(-16\right)\hfill \\ \text{ }-10x - 15y=80\hfill \\ \text{ }2\left(5x - 10y\right)=2\left(30\right)\hfill \\ \text{ }10x - 20y=60\hfill \end{array}[/latex-display] Then, we add the two equations together. [latex-display]\begin{array}\ −10x−15y=80 \\ 10x−20y=60 \\ \text{______________} \\ \text{ }−35y=140 \\ y=−4 \end{array}[/latex-display] Substitute $y=-4$ into the original first equation. [latex-display]\begin{array}{c}2x+3\left(-4\right)=-16\\ 2x - 12=-16\\ 2x=-4\\ x=-2\end{array}[/latex-display] The solution is $\left(-2,-4\right)$. Check it in the other equation. [latex-display]\begin{array}{r}\hfill \text{ }5x - 10y=30\\ \hfill 5\left(-2\right)-10\left(-4\right)=30\\ \hfill \text{ }-10+40=30\\ \hfill \text{ }30=30\end{array}[/latex-display]

### Example: Using the Addition Method in Systems of Equations Containing Fractions

Solve the given system of equations in two variables by addition. [latex-display]\begin{array}{l}\frac{x}{3}+\frac{y}{6}=3\hfill \\ \frac{x}{2}-\frac{y}{4}=\text{ }1\hfill \end{array}[/latex-display]

Answer: First clear each equation of fractions by multiplying both sides of the equation by the least common denominator. [latex-display]\begin{array}{l}6\left(\frac{x}{3}+\frac{y}{6}\right)=6\left(3\right)\hfill \\ \text{ }2x+y=18\hfill \\ 4\left(\frac{x}{2}-\frac{y}{4}\right)=4\left(1\right)\hfill \\ \text{ }2x-y=4\hfill \end{array}[/latex-display] Now multiply the second equation by $-1$ so that we can eliminate the x-variable. [latex-display]\begin{array}{l}-1\left(2x-y\right)=-1\left(4\right)\hfill \\ \text{ }-2x+y=-4\hfill \end{array}[/latex-display] Add the two equations to eliminate the x-variable and solve the resulting equation. [latex-display]\begin{array}\ \hfill 2x+y=18 \\ \hfill−2x+y=−4 \\ \text{_____________} \\ \hfill 2y=14 \\ \hfill y=7 \end{array}[/latex-display] Substitute $y=7$ into the first equation. [latex-display]\begin{array}{l}2x+\left(7\right)=18\hfill \\ \text{ }2x=11\hfill \\ \text{ }x=\frac{11}{2}\hfill \\ \text{ }=7.5\hfill \end{array}[/latex-display] The solution is $\left(\frac{11}{2},7\right)$. Check it in the other equation. [latex-display]\begin{array}{c}\frac{x}{2}-\frac{y}{4}=1\\ \frac{\frac{11}{2}}{2}-\frac{7}{4}=1\\ \frac{11}{4}-\frac{7}{4}=1\\ \frac{4}{4}=1\end{array}[/latex-display]

### Try It

Solve the system of equations by addition. [latex-display]\begin{array}{c}2x+3y=8\\ 3x+5y=10\end{array}[/latex-display]

Answer: $\left(10,-4\right)$

### Expressing the Solution of a System of Dependent Equations Containing Two Variables

Recall that a dependent system of equations in two variables is a system in which the two equations represent the same line. Dependent systems have an infinite number of solutions because all of the points on one line are also on the other line. After using substitution or addition, the resulting equation will be an identity, such as $0=0$.

### Example: Finding a Solution to a Dependent System of Linear Equations

Find a solution to the system of equations using the addition method.

$\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}$

Answer: With the addition method, we want to eliminate one of the variables by adding the equations. In this case, let’s focus on eliminating $x$. If we multiply both sides of the first equation by $-3$, then we will be able to eliminate the $x$ -variable.

$\begin{array}{l}\text{ }x+3y=2\hfill \\ \left(-3\right)\left(x+3y\right)=\left(-3\right)\left(2\right)\hfill \\ \text{ }-3x - 9y=-6\hfill \end{array}$

$\begin{array} \hfill−3x−9y=−6 \\ \hfill+3x+9y=6 \\ \hfill \text{_____________} \\ \hfill 0=0 \end{array}$

We can see that there will be an infinite number of solutions that satisfy both equations.

#### Analysis of the Solution

If we rewrote both equations in the slope-intercept form, we might know what the solution would look like before adding. Let’s look at what happens when we convert the system to slope-intercept form.

$\begin{array}{l}\text{ }x+3y=2\hfill \\ \text{ }3y=-x+2\hfill \\ \text{ }y=-\frac{1}{3}x+\frac{2}{3}\hfill \\ 3x+9y=6\hfill \\ \text{ }9y=-3x+6\hfill \\ \text{ }y=-\frac{3}{9}x+\frac{6}{9}\hfill \\ \text{ }y=-\frac{1}{3}x+\frac{2}{3}\hfill \end{array}$

Look at the graph below. Notice the results are the same. The general solution to the system is $\left(x, -\frac{1}{3}x+\frac{2}{3}\right)$.

### Writing the general solution

In the previous example, we presented an analysis of the solution to the following system of equations:

$\begin{array}{c}x+3y=2\\ 3x+9y=6\end{array}$

After a little algebra, we found that these two equations were exactly the same. We then wrote the general solution as $\left(x, -\frac{1}{3}x+\frac{2}{3}\right)$. Why would we write the solution this way? In some ways, this representation tells us a lot.  It tells us that x can be anything, x is x.  It also tells us that y is going to depend on x, just like when we write a function rule.  In this case, depending on what you put in for x, y will be defined in terms of x as $-\frac{1}{3}x+\frac{2}{3}$. In other words, there are infinitely many (x,y) pairs that will satisfy this system of equations, and they all fall on the line $f(x)-\frac{1}{3}x+\frac{2}{3}$.

### Try It

Solve the following system of equations in two variables.

$\begin{array}{l}\begin{array}{l}\\ \text{ }\text{}\text{}y - 2x=5\end{array}\hfill \\ -3y+6x=-15\hfill \end{array}$

Answer: The system is dependent so there are infinite solutions of the form $\left(x,2x+5\right)$.

Sometimes, a system can inform a decision.  In our next example, we help answer the question, "Which truck rental company will give the best value?"

### Example: Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of$16, then 63 cents a mile.[footnote]Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/[/footnote] When will Keep on Trucking, Inc. be the better choice for Jamal?

Answer: The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

 Input d, distance driven in miles Outputs K(d): cost, in dollars, for renting from Keep on TruckingM(d) cost, in dollars, for renting from Move It Your Way Initial Value Up-front fee: K(0) = 20 and M(0) = 16 Rate of Change K(d) = $0.59/mile and P(d) =$0.63/mile
A linear function is of the form $f\left(x\right)=mx+b$. Using the rates of change and initial charges, we can write the equations

$\begin{array}{l}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{array}$

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when $K\left(d\right)<M\left(d\right)$. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the $K\left(d\right)$ function is smaller. These graphs are sketched above, with K(d) in blue. To find the intersection, we set the equations equal and solve:

$\begin{array}{l}K\left(d\right)=M\left(d\right)\hfill \\ 0.59d+20=0.63d+16\hfill \\ 4=0.04d\hfill \\ 100=d\hfill \\ d=100\hfill \end{array}$

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that $K\left(d\right)$ is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is $d>100$.

The applications for systems seems almost endless, but we will just show one more. In the next example, we determine the amount 80% methane solution to add to a 50% solution to give a final solution of 60%.

### Example: Solve a Chemical Mixture Problem

A chemist has 70 mL of a 50% methane solution. How much of a 80% solution must she add so the final solution is 60% methane?

Answer: We will use the following table to help us solve this mixture problem:

 Amount Part Total Start Add Final
We start with 70 mL of solution, and the unknown amount can be x. The part is the percentages, or concentration of solution 0.5 for start, 0.8 for add.
 Amount Part Total Start 70mL 0.5 Add x 0.8 Final $70+x$ 0.6
Add amount column to get final amount. The part for this amount is 0.6 because we want the final solution to be 60% methane.
 Amount Part Total Start 70mL 0.5 35 Add x 0.8 $0.8x$ Final $70+x$ 0.6 $42+0.6x$
Multiply amount by part to get total. be sure to distribute on the last row:$(70 + x)0.6$. If we add the start and add entries in the Total column, we get the final equation that represents the total amount and it's concentration. [latex-display]\begin{array}{c}35+0.8x = 42+0.6x\\0.2x=7\\\frac{0.2}{0.2}x=\frac{7}{0.2}\\x=35\end{array}[/latex-display] 35mL of 80% solution must be added to 70mL of 50% solution to get a 60% solution of Methane. The same process can be used if the starting and final amount have a price attached to them, rather than a percentage.

## Key Concepts

• A system of linear equations consists of two or more equations made up of two or more variables such that all equations in the system are considered simultaneously.
• The solution to a system of linear equations in two variables is any ordered pair that satisfies each equation independently.
• Systems of equations are classified as independent with one solution, dependent with an infinite number of solutions, or inconsistent with no solution.
• One method of solving a system of linear equations in two variables is by graphing. In this method, we graph the equations on the same set of axes.
• Another method of solving a system of linear equations is by substitution. In this method, we solve for one variable in one equation and substitute the result into the second equation.
• A third method of solving a system of linear equations is by addition, in which we can eliminate a variable by adding opposite coefficients of corresponding variables.
• It is often necessary to multiply one or both equations by a constant to facilitate elimination of a variable when adding the two equations together.
• Either method of solving a system of equations results in a false statement for inconsistent systems because they are made up of parallel lines that never intersect.
• The solution to a system of dependent equations will always be true because both equations describe the same line.
• Systems of equations can be used to solve real-world problems that involve more than one variable, such as those relating to revenue, cost, and profit.

## Glossary

addition method an algebraic technique used to solve systems of linear equations in which the equations are added in a way that eliminates one variable, allowing the resulting equation to be solved for the remaining variable; substitution is then used to solve for the first variable break-even point the point at which a cost function intersects a revenue function; where profit is zero consistent system a system for which there is a single solution to all equations in the system and it is an independent system, or if there are an infinite number of solutions and it is a dependent system cost function the function used to calculate the costs of doing business; it usually has two parts, fixed costs and variable costs dependent system a system of linear equations in which the two equations represent the same line; there are an infinite number of solutions to a dependent system inconsistent system a system of linear equations with no common solution because they represent parallel lines, which have no point or line in common independent system a system of linear equations with exactly one solution pair $\left(x,y\right)$ profit function the profit function is written as $P\left(x\right)=R\left(x\right)-C\left(x\right)$, revenue minus cost revenue function the function that is used to calculate revenue, simply written as $R=xp$, where $x=$ quantity and $p=$ price substitution method an algebraic technique used to solve systems of linear equations in which one of the two equations is solved for one variable and then substituted into the second equation to solve for the second variable system of linear equations a set of two or more equations in two or more variables that must be considered simultaneously.