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## How to Solve Systems of Algebraic Equations Containing Two Variables

Last Updated: July 30, 2023 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. There are 8 references cited in this article, which can be found at the bottom of the page. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 1,027,843 times.

In a "system of equations," you are asked to solve two or more equations at the same time. When these have two different variables in them, such as x and y, or a and b, it can be tricky at first glance to see how to solve them. [1] X Research source Fortunately, once you know what to do, all you need is basic algebra skills (and sometimes some knowledge of fractions) to solve the problem. If you are a visual learner or if your teacher requires it, learn how to graph the equations as well. Graphing can be useful to "see what's going on" or to check your work, but it can be slower than the other methods, and doesn't work well for all systems of equations.

## Using the Substitution Method

• This method often uses fractions later on. You can try the elimination method below instead if you don't like fractions.

• 4x = 8 - 2y
• (4x)/4 = (8/4) - (2y/4)

• You know that x = 2 - ½y .
• Your second equation, that you haven't yet altered, is 5x + 3y = 9 .
• In the second equation, replace x with "2 - ½y": 5(2 - ½y) + 3y = 9 .

• 5(2 - ½y) + 3y = 9
• 10 – (5/2)y + 3y = 9
• 10 – (5/2)y + (6/2)y = 9 (If you don't understand this step, learn how to add fractions . This is often, but not always, necessary for this method.)
• 10 + ½y = 9

• You know that y = -2
• One of the original equations is 4x + 2y = 8 . (You can use either equation for this step.)
• Plug in -2 instead of y: 4x + 2(-2) = 8 .

• If you end up with an equation that has no variables and isn't true (for instance, 3 = 5), the problem has no solution . (If you graphed both of the equations, you'd see they were parallel and never intersect.)
• If you end up with an equation without variables that is true (such as 3 = 3), the problem has infinite solutions . The two equations are exactly equal to each other. (If you graphed the two equations, you'd see they were the same line.)

## Using the Elimination Method

• You have the system of equations 3x - y = 3 and -x + 2y = 4 .
• Let's change the first equation so that the y variable will cancel out. (You can choose x instead, and you'll get the same answer in the end.)
• The - y on the first equation needs to cancel with the + 2y in the second equation. We can make this happen by multiplying - y by 2.
• Multiply both sides of the first equation by 2, like this: 2(3x - y)=2(3) , so 6x - 2y = 6 . Now the - 2y will cancel out with the +2y in the second equation.

• Your equations are 6x - 2y = 6 and -x + 2y = 4 .
• Combine the left sides: 6x - 2y - x + 2y = ?
• Combine the right sides: 6x - 2y - x + 2y = 6 + 4 .

• You have 6x - 2y - x + 2y = 6 + 4 .
• Group the x and y variables together: 6x - x - 2y + 2y = 6 + 4 .
• Simplify: 5x = 10
• Solve for x: (5x)/5 = 10/5 , so x = 2 .

• You know that x = 2 , and one of your original equations is 3x - y = 3 .
• Plug in 2 instead of x: 3(2) - y = 3 .
• Solve for y in the equation: 6 - y = 3
• 6 - y + y = 3 + y , so 6 = 3 + y

• If your combined equation has no variables and is not true (like 2 = 7), there is no solution that will work on both equations. (If you graph both equations, you'll see they're parallel and never cross.)
• If your combined equation has no variables and is true (like 0 = 0), there are infinite solutions . The two equations are actually identical. (If you graph them, you'll see that they're the same line.)

## Graphing the Equations

• The basic idea is to graph both equations, and find the point where they intersect. The x and y values at this point will give us the value of x and the value of y in the system of equations.

• Your first equation is 2x + y = 5 . Change this to y = -2x + 5 .
• Your second equation is -3x + 6y = 0 . Change this to 6y = 3x + 0 , then simplify to y = ½x + 0 .
• If both equations are identical , the entire line will be an "intersection". Write infinite solutions .

• If you don't have graph paper, use a ruler to make sure the numbers are spaced precisely apart.
• If you are using large numbers or decimals, you may need to scale your graph differently. (For example, 10, 20, 30 or 0.1, 0.2, 0.3 instead of 1, 2, 3).

• In our examples from earlier, one line ( y = -2x + 5 ) intercepts the y-axis at 5 . The other ( y = ½x + 0 ) intercepts at 0 . (These are points (0,5) and (0,0) on the graph.)
• Use different colored pens or pencils if possible for the two lines.

• In our example, the line y = -2x + 5 has a slope of -2 . At x = 1, the line moves down 2 from the point at x = 0. Draw the line segment between (0,5) and (1,3).
• The line y = ½x + 0 has a slope of ½ . At x = 1, the line moves up ½ from the point at x=0. Draw the line segment between (0,0) and (1,½).
• If the lines have the same slope , the lines will never intersect, so there is no answer to the system of equations. Write no solution .

• If the lines are moving toward each other, keep plotting points in that direction.
• If the lines are moving away from each other, move back and plot points in the other direction, starting at x = -1.
• If the lines are nowhere near each other, try jumping ahead and plotting more distant points, such as at x = 10.

## Video . By using this service, some information may be shared with YouTube.

• You can check your work by plugging the answers back into the original equations. If the equations end up true (for instance, 3 = 3), your answer is correct. Thanks Helpful 3 Not Helpful 1
• In the elimination method, you will sometimes have to multiply one equation by a negative number in order to get a variable to cancel out. Thanks Helpful 1 Not Helpful 1

• These methods cannot be used if there is a variable raised to an exponent, such as x 2 . For more information on equations of this type, look up a guide to factoring quadratics with two variables. [11] X Research source Thanks Helpful 0 Not Helpful 0

## You Might Also Like

• ↑ https://www.mathsisfun.com/definitions/system-of-equations.html
• ↑ https://calcworkshop.com/systems-equations/substitution-method/
• ↑ https://www.cuemath.com/algebra/substitution-method/
• ↑ https://tutorial.math.lamar.edu/Classes/Alg/SystemsTwoVrble.aspx
• ↑ http://www.purplemath.com/modules/systlin2.htm
• ↑ http://www.virtualnerd.com/algebra-2/linear-systems/graphing/solve-by-graphing/equations-solution-by-graphing

To solve systems of algebraic equations containing two variables, start by moving the variables to different sides of the equation. Then, divide both sides of the equation by one of the variables to solve for that variable. Next, take that number and plug it into the formula to solve for the other variable. Finally, take your answer and plug it into the original equation to solve for the other variable. To learn how to solve systems of algebraic equations using the elimination method, scroll down! Did this summary help you? Yes No

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## Course: Algebra 1   >   Unit 4

• Two-variable linear equations intro

## Solutions to 2-variable equations

• Worked example: solutions to 2-variable equations
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## How Do You Solve Two Equations with Two Variables?

Trying to solve two equations each with the same two unknown variables? Take one of the equations and solve it for one of the variables. Then plug that into the other equation and solve for the variable. Plug that value into either equation to get the value for the other variable. This tutorial will take you through this process of substitution step-by-step!

• substitution
• solve by substitution
• apply distributive property
• linear equation
• 2 equations 2 unknowns

## Background Tutorials

Evaluating expressions.

## What is a Variable?

You can't do algebra without working with variables, but variables can be confusing. If you've ever wondered what variables are, then this tutorial is for you!

## How Do You Use the Distributive Property to Simplify an Expression?

In this tutorial you'll see how to apply the distributive property. Remember that this is important when you are trying to simplify an expression and get rid of parentheses!

## How Do You Simplify an Expression?

Simplifying an algebraic expression is a fundamental part of solving math problems. Get some practice putting an expression in simplest form by following along with this tutorial.

## How Do You Solve a Two-Step Equation?

Solving an equation for a variable? Perform the order of operations in reverse! Check it out in this tutorial.

## How Do You Check Your Answer When You Have Two Equations?

Imagine you have two equations with two variables that you're trying to solve for, and someone hands you the answer. How do you know that the answer is right? After watching this tutorial you'll see exactly what it takes to check that the answer you have is correct for BOTH equations!

## Further Exploration

Solving systems using substitution.

## How Do You Solve a Word Problem Using Two Equations?

Sometimes word problems describe a system of equations, two equations each with two unknowns. Solving word problems like this one aren't so bad if you know what to do. Check it out with this tutorial!

## What Are the Ways You Can Solve a System of Linear Equations?

Knowing the definition of a system of equations is great, but you should also know how to solve them! This tutorial introduces you to the graphing method, substitution method, and elimination method for solving a system of equations. Take a look and learn them all!

## Solving Equations With Two Variables

Related Pages Solving Equations More Lessons for GRE Math More Algebra Lessons

This is part of a series of lessons for the quantitative reasoning section of the GRE revised General Test. In these lessons, we will learn:

• Linear Equations in Two Variables
• Solving Simultaneous Equations or Systems of Equations
• Using the Substitution Method
• Using the Elimination Method

## Linear Equation In Two Variables

A linear equation in two variables, x and y, can be written in the form ax + by = c where x and y are real numbers and a and b are not both zero.

For example, 3x + 2y = 8 is a linear equation in two variables.

A solution of such an equation is an ordered pair of numbers (x, y) that makes the equation true when the values of x and y are substituted into the equation.

For example, both (2, 1) and (0, 4) are solutions of the equation but (2, 0) is not a solution. A linear equation in two variables has infinitely many solutions.

The following video shows how to complete ordered pairs to make a solution to linear equations.

## Systems of Equations or Simultaneous Equations

The following diagram shows examples of how to solve systems of equations using substitution or elimination.. Scroll down the page for more examples and solutions on how to solve systems of equations or simultaneous equations..

If another linear equation in the same variables is given, it is usually possible to find a unique solution of both equations. Two equations with the same variables are called a system of equations , and the equations in the system are called simultaneous equations . To solve a system of two equations means to find an ordered pair of numbers that satisfies both equations in the system.

There are two basic methods for solving systems of linear equations, by substitution or by elimination.

## Substitution Method

In the substitution method, one equation is manipulated to express one variable in terms of the other. Then the expression is substituted in the other equation.

For example, to solve the system of equations 3x + 2y = 2 y + 8 = 3x

Isolate the variable y in the equation y + 8 = 3x to get y = 3x – 8.

Then, substitute 3x – 8 for y into the equation 3x + 2y = 2. 3x + 2 (3x – 8) = 2 3x + 6x – 16 = 2 9x – 16 = 2 9x = 18

Substitute x = 2 into y = 3x – 8.to get the value for y y = 3 (2) – 8 y = 6 – 8 = – 2

Answer: x = 2 and y = –2

How to solve simultaneous equations using substitution?

## Elimination Method

In the elimination method, the object is to make the coefficients of one variable the same in both equations so that one variable can be eliminated either by adding the equations together or by subtracting one from the other.

Consider the following example: 2x + 3y = –2 4x – 3y = 14

In this example the coefficients of y are already opposites (+3 and –3). Just add the two equations to eliminate y.

To get the value of y, we need to substitute x = 2 into the equation 2x + 3y = –2 2(2) + 3y = –2 4 + 3y = –2 3y = –6 y = –2

How to solve simultaneous equations using the substitution method and elimination (or combination) method

Example of the GRE Quantitative Comparison question that involves simultaneous equations

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• Quiz: Square Roots and Cube Roots
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• Quiz: Signed Numbers (Positive Numbers and Negative Numbers)
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Here are some examples for solving number problems with two variables.

The sum of two numbers is 15. The difference of the same two numbers is 7. What are the two numbers?

First, circle what you're looking for— the two numbers. Let x stand for the larger number and y stand for the second number. Now, set up two equations.

The sum of the two numbers is 15.

The difference is 7.

Now, solve by adding the two equations.

Now, plugging into the first equation gives

The numbers are 11 and 4.

The sum of twice one number and three times another number is 23 and their product is 20. Find the numbers.

First, circle what you must find— the numbers . Let x stand for the number that is being multiplied by 2 and y stand for the number being multiplied by 3.

Now set up two equations.

The sum of twice a number and three times another number is 23.

2 x + 3 y = 23

Their product is 20.

x ( y ) = 20

Rearranging the first equation gives

3 y = 23 – 2 x

Dividing each side of the equation by 3 gives

Now, substituting the first equation into the second gives

Multiplying each side of the equation by 3 gives

23 x – 2 x 2 = 60

Rewriting this equation in standard quadratic form gives

2 x 2 – 23 x + 60 = 0

Solving this quadratic equation using factoring gives

(2 x – 15)( x – 4) = 0

Setting each factor equal to 0 and solving gives

With each x value we can find its corresponding y value.

Therefore, this problem has two sets of solutions.

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## Module 13: Systems of Equations and Inequalities

Systems of linear equations: two variables, learning outcomes.

• Solve systems of equations by graphing, substitution, and addition.
• Identify inconsistent systems of equations containing two variables.
• Express the solution of a system of dependent equations containing two variables using standard notations.

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{align}2x+y&=15\\[1mm] 3x-y&=5\end{align}

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{align}2\left(4\right)+\left(7\right)&=15 &&\text{True} \\[1mm] 3\left(4\right)-\left(7\right)&=5 &&\text{True} \end{align}

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.

• Substitute the ordered pair into each equation in the system.
• 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{align}x+3y&=8\\ 2x-9&=y \end{align}

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

\begin{align}\left(5\right)+3\left(1\right)&=8 \\[1mm] 8&=8 &&\text{True} \\[3mm] 2\left(5\right)-9&=\left(1\right) \\[1mm] 1&=1 &&\text{True} \end{align}

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.

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

\begin{align}5x-4y&=20\\ 2x+1&=3y\end{align}

Not a solution.

## 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{align}2x+y&=-8\\ x-y&=-1\end{align}

Solve the first equation for $y$.

\begin{align}2x+y&=-8\\ y&=-2x-8\end{align}

Solve the second equation for $y$.

\begin{align}x-y&=-1\\ y&=x+1\end{align}

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{align}2\left(-3\right)+\left(-2\right)&=-8 \\[1mm] -8=-8 &&\text{True} \\[3mm] \left(-3\right)-\left(-2\right)&=-1 \\[1mm] -1&=-1 &&\text{True} \end{align}

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

Solve the following system of equations by graphing.

$\begin{gathered}2x - 5y=-25 \\ -4x+5y=35 \end{gathered}$

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

## 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.

Plot the three different systems with an online graphing tool. Categorize each solution as either consistent or inconsistent. If the system is consistent determine whether it is dependent or independent. You may find it easier to plot each system individually, then clear out your entries before you plot the next. 1) $5x-3y = -19$ $x=2y-1$

2) $4x+y=11$ $-2y=-25+8x$

3) $y = -3x+6$ $-\frac{1}{3}y+2=x$

• One solution – consistent, independent
• No solutions, inconsistent, neither dependent nor independent
• Many solutions –  consistent, dependent

## 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.

• Solve one of the two equations for one of the variables in terms of the other.
• Substitute the expression for this variable into the second equation, then solve for the remaining variable.
• 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.
• 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.

\begin{align}-x+y&=-5 \\ 2x-5y&=1 \end{align}

First, we will solve the first equation for $y$.

\begin{align}-x+y&=-5 \\ y&=x - 5 \end{align}

Now we can substitute the expression $x - 5$ for $y$ in the second equation.

\begin{align}2x - 5y&=1 \\ 2x - 5\left(x - 5\right)&=1 \\ 2x - 5x+25&=1 \\ -3x&=-24 \\ x&=8 \end{align}

Now, we substitute $x=8$ into the first equation and solve for $y$.

\begin{align}-\left(8\right)+y&=-5 \\ y&=3 \end{align}

Our solution is $\left(8,3\right)$.

Check the solution by substituting $\left(8,3\right)$ into both equations.

\begin{align}-x+y&=-5 \\ -\left(8\right)+\left(3\right)&=-5 && \text{True} \\[3mm] 2x - 5y&=1 \\ 2\left(8\right)-5\left(3\right)&=1 && \text{True} \end{align}

\begin{align}x&=y+3 \\ 4&=3x - 2y \end{align}

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

## Can the substitution method be used to solve any linear system in two variables?

Yes, but the method works best if one of the equations contains a coefficient of 1 or –1 so that we do not have to deal with fractions.

The following video is ~10 minutes long and provides a mini-lesson on using the substitution method to solve a system of linear equations.  We present three different examples, and also use a graphing tool to help summarize the solution for each example.

## Solving Systems of Equations in Two Variables by the Addition Method

A third method of solving systems of linear equations is the addition method,  this method is also called the  elimination method . In this method, we add two terms with the same variable, but opposite coefficients, so that the sum is zero. Of course, not all systems are set up with the two terms of one variable having opposite coefficients. Often we must adjust one or both of the equations by multiplication so that one variable will be eliminated by addition.

## How To: Given a system of equations, solve using the addition method.

• Write both equations with x – and y -variables on the left side of the equal sign and constants on the right.
• Write one equation above the other, lining up corresponding variables. If one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, add the equations together, eliminating one variable. If not, use multiplication by a nonzero number so that one of the variables in the top equation has the opposite coefficient of the same variable in the bottom equation, then add the equations to eliminate the variable.
• Solve the resulting equation for the remaining variable.
• Substitute that value into one of the original equations and solve for the second variable.
• Check the solution by substituting the values into the other equation.

## Example: Solving a System by the Addition Method

Solve the given system of equations by addition.

\begin{align}x+2y&=-1 \\ -x+y&=3 \end{align}

Both equations are already set equal to a constant. Notice that the coefficient of $x$ in the second equation, –1, is the opposite of the coefficient of $x$ in the first equation, 1. We can add the two equations to eliminate $x$ without needing to multiply by a constant.

\begin{align} x+2y&=-1 \\ -x+y&=3 \\ \hline 3y&=2\end{align}

Now that we have eliminated $x$, we can solve the resulting equation for $y$.

\begin{align}3y&=2 \\ y&=\dfrac{2}{3} \end{align}

Then, we substitute this value for $y$ into one of the original equations and solve for $x$.

\begin{align}-x+y&=3 \\ -x+\frac{2}{3}&=3 \\ -x&=3-\frac{2}{3} \\ -x&=\frac{7}{3} \\ x&=-\frac{7}{3} \end{align}

The solution to this system is $\left(-\frac{7}{3},\frac{2}{3}\right)$.

Check the solution in the first equation.

\begin{align}x+2y&=-1 \\ \left(-\frac{7}{3}\right)+2\left(\frac{2}{3}\right)&= \\ -\frac{7}{3}+\frac{4}{3}&= \\ \-\frac{3}{3}&= \\ -1&=-1&& \text{True} \end{align}

We gain an important perspective on systems of equations by looking at the graphical representation. See the graph below to find that the equations intersect at the solution. We do not need to ask whether there may be a second solution because observing the graph confirms that the system has exactly one solution.

## Example: Using the Addition Method When Multiplication of One Equation Is Required

Solve the given system of equations by the addition method .

\begin{align}3x+5y&=-11 \\ x - 2y&=11 \end{align}

Adding these equations as presented will not eliminate a variable. However, we see that the first equation has $3x$ in it and the second equation has $x$. So if we multiply the second equation by $-3,\text{}$ the x -terms will add to zero.

\begin{align}x - 2y&=11 \\ -3\left(x - 2y\right)&=-3\left(11\right) && \text{Multiply both sides by }-3 \\ -3x+6y&=-33 && \text{Use the distributive property}. \end{align}

\begin{align}3x+5y&=−11 \\ −3x+6y&=−33 \\ \hline 11y&=−44 \\ y&=−4 \end{align}

For the last step, we substitute $y=-4$ into one of the original equations and solve for $x$.

\begin{align}3x+5y&=-11\\ 3x+5\left(-4\right)&=-11\\ 3x - 20&=-11\\ 3x&=9\\ x&=3\end{align}

Our solution is the ordered pair $\left(3,-4\right)$. Check the solution in the original second equation.

\begin{align}x - 2y&=11 \\ \left(3\right)-2\left(-4\right)&=3+8 \\ &=11 && \text{True} \end{align}

Solve the system of equations by addition.

\begin{align}2x - 7y&=2\\ 3x+y&=-20\end{align}

$\left(-6,-2\right)$

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

Solve the given system of equations in two variables by addition.

\begin{align}2x+3y&=-16 \\ 5x - 10y&=30\end{align}

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 $2$.

\begin{align} -5\left(2x+3y\right)&=-5\left(-16\right) \\ -10x - 15y&=80 \\[3mm] 2\left(5x - 10y\right)&=2\left(30\right) \\ 10x - 20y&=60 \end{align}

Then, we add the two equations together.

\begin{align} −10x−15y&=80 \\ 10x−20y&=60 \\ \hline −35y&=140 \\ y&=−4 \end{align}

Substitute $y=-4$ into the original first equation.

\begin{align}2x+3\left(-4\right)&=-16\\ 2x - 12&=-16\\ 2x&=-4\\ x&=-2\end{align}

The solution is $\left(-2,-4\right)$. Check it in the other equation.

\begin{align} 5x - 10y&=30\\ 5\left(-2\right)-10\left(-4\right)&=30\\ -10+40&=30\\ 30&=30\end{align}

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

\begin{align}\frac{x}{3}+\frac{y}{6}&=3 \\[1mm] \frac{x}{2}-\frac{y}{4}&=1 \end{align}

First clear each equation of fractions by multiplying both sides of the equation by the least common denominator.

\begin{align}6\left(\frac{x}{3}+\frac{y}{6}\right)&=6\left(3\right) \\[1mm] 2x+y&=18 \\[3mm] 4\left(\frac{x}{2}-\frac{y}{4}\right)&=4\left(1\right) \\[1mm] 2x-y&=4 \end{align}

Now multiply the second equation by $-1$ so that we can eliminate  x .

\begin{align}-1\left(2x-y\right)&=-1\left(4\right) \\[1mm] -2x+y&=-4 \end{align}

Add the two equations to eliminate  x  and solve the resulting equation for y .

\begin{align} 2x+y&=18 \\ −2x+y&=−4 \\ \hline 2y&=14 \\ y&=7 \end{align}

Substitute $y=7$ into the first equation.

\begin{align}2x+\left(7\right)&=18 \\ 2x&=11 \\ x&=\frac{11}{2} \\ &=7.5 \end{align}

The solution is $\left(\frac{11}{2},7\right)$. Check it in the other equation.

\begin{align}\frac{x}{2}-\frac{y}{4}&=1\\[1mm] \frac{\frac{11}{2}}{2}-\frac{7}{4}&=1\\[1mm] \frac{11}{4}-\frac{7}{4}&=1\\[1mm] \frac{4}{4}&=1\end{align}

\begin{align}2x+3y&=8\\ 3x+5y&=10\end{align}

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

in the following video we present more examples of how to use the addition (elimination) method to solve a system of two linear equations.

## Classify Solutions to Systems

Now that we have several methods for solving systems of equations, we can use the methods to identify inconsistent systems. Recall that an inconsistent system consists of parallel lines that have the same slope but different $y$ -intercepts. They will never intersect. When searching for a solution to an inconsistent system, we will come up with a false statement, such as $12=0$.

## Example: Solving an Inconsistent System of Equations

Solve the following system of equations.

$\begin{gathered}&x=9 - 2y \\ &x+2y=13 \end{gathered}$

We can approach this problem in two ways. Because one equation is already solved for $x$, the most obvious step is to use substitution.

\begin{align}x+2y&=13 \\ \left(9 - 2y\right)+2y&=13 \\ 9+0y&=13 \\ 9&=13 \end{align}

Clearly, this statement is a contradiction because $9\ne 13$. Therefore, the system has no solution.

The second approach would be to first manipulate the equations so that they are both in slope-intercept form. We manipulate the first equation as follows.

$\begin{gathered}x=9 - 2y \\ 2y=-x+9 \\ y=-\frac{1}{2}x+\frac{9}{2} \end{gathered}$

We then convert the second equation expressed to slope-intercept form.

$\begin{gathered}x+2y=13 \\ 2y=-x+13 \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}$

Comparing the equations, we see that they have the same slope but different y -intercepts. Therefore, the lines are parallel and do not intersect.

$\begin{gathered}y=-\frac{1}{2}x+\frac{9}{2} \\ y=-\frac{1}{2}x+\frac{13}{2} \end{gathered}$

Writing the equations in slope-intercept form confirms that the system is inconsistent because all lines will intersect eventually unless they are parallel. Parallel lines will never intersect; thus, the two lines have no points in common. The graphs of the equations in this example are shown below.

Solve the following system of equations in two variables.

$\begin{gathered}2y - 2x=2\\ 2y - 2x=6\end{gathered}$

No solution. It is an inconsistent system.

## 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 .

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

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{align}x+3y&=2 \\ \left(-3\right)\left(x+3y\right)&=\left(-3\right)\left(2\right) \\ -3x - 9y&=-6 \end{align}

\begin{align} −3x−9y&=−6 \\ +3x+9y&=6 \\ \hline 0&=0 \end{align}

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

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{align}\begin{gathered}x+3y=2 \\ 3y=-x+2 \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered} \hspace{2cm} \begin{gathered} 3x+9y=6 \\9y=-3x+6 \\ y=-\frac{3}{9}x+\frac{6}{9} \\ y=-\frac{1}{3}x+\frac{2}{3} \end{gathered}\end{align}

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:

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 $f(x)-\frac{1}{3}x+\frac{2}{3}$.

$\begin{gathered}y - 2x=5 \\ -3y+6x=-15 \end{gathered}$

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

## Using Systems of Equations to Investigate Profits

Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation $R=xp$, where $x=$ quantity and $p=$ price. The revenue function is shown in orange in the graph below.

The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The x -axis represents quantity in hundreds of units. The y -axis represents either cost or revenue in hundreds of dollars.

The point at which the two lines intersect is called the break-even point . We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also$3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as $P\left(x\right)=R\left(x\right)-C\left(x\right)$. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

## Example: Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function $C\left(x\right)=0.85x+35{,}000$ and the revenue function $R\left(x\right)=1.55x$, find the break-even point and the profit function.

Write the system of equations using $y$ to replace function notation.

\begin{align} y&=0.85x+35{,}000 \\ y&=1.55x \end{align}

Substitute the expression $0.85x+35{,}000$ from the first equation into the second equation and solve for $x$.

$\begin{gathered}0.85x+35{,}000=1.55x\\ 35{,}000=0.7x\\ 50{,}000=x\end{gathered}$

Then, we substitute $x=50{,}000$ into either the cost function or the revenue function.

$1.55\left(50{,}000\right)=77{,}500$

The break-even point is $\left(50{,}000,77{,}500\right)$.

The profit function is found using the formula $P\left(x\right)=R\left(x\right)-C\left(x\right)$.

\begin{align}P\left(x\right)&=1.55x-\left(0.85x+35{,}000\right) \\ &=0.7x - 35{,}000 \end{align}

The profit function is $P\left(x\right)=0.7x - 35{,}000$.

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also$77,500. To make a profit, the business must produce and sell more than 50,000 units.

We see from the graph below that the profit function has a negative value until $x=50{,}000$, when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

## Writing a System of Linear Equations Given a Situation

It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.

## How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

• Identify the input and output of each linear model.
• Identify the slope and y -intercept of each linear model.
• Find the solution by setting the two linear functions equal to another and solving for x , or find the point of intersection on a graph.

Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.

## Example: Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $25.00 for children and$50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets? Let c = the number of children and a = the number of adults in attendance. The total number of people is 2,000. We can use this to write an equation for the number of people at the circus that day. $c+a=2{,}000$ The revenue from all children can be found by multiplying$25.00 by the number of children, $25c$. The revenue from all adults can be found by multiplying $50.00 by the number of adults, $50a$. The total revenue is$70,000. We can use this to write an equation for the revenue.

$25c+50a=70{,}000$

We now have a system of linear equations in two variables.

$\begin{gathered}c+a=2,000\\ 25c+50a=70{,}000\end{gathered}$

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either $c$ or $a$. We will solve for $a$.

$\begin{gathered}c+a=2{,}000\\ a=2{,}000-c\end{gathered}$

Substitute the expression $2{,}000-c$ in the second equation for $a$ and solve for $c$.

\begin{align} 25c+50\left(2{,}000-c\right)&=70{,}000 \\ 25c+100{,}000 - 50c&=70{,}000 \\ -25c&=-30{,}000 \\ c&=1{,}200 \end{align}

Substitute $c=1{,}200$ into the first equation to solve for $a$.

\begin{align}1{,}200+a&=2{,}000 \\ a&=800 \end{align}

We find that 1,200 children and 800 adults bought tickets to the circus that day.

Meal tickets at the circus cost $4.00 for children and$12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets? 700 children, 950 adults Sometimes, a system of equations 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. [1] When will Keep on Trucking, Inc. be the better choice for Jamal?

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.

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{align}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{align}

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{align}K\left(d\right)&=M\left(d\right) \\ 0.59d+20&=0.63d+16 \\ 4&=0.04d \\ 100&=d \\ d&=100 \end{align}

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?

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

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.

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.

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.

\begin{align}35+0.8x& = 42+0.6x \\ 0.2x&=7 \\ \frac{0.2}{0.2}x&=\frac{7}{0.2} \\ x&=35 \end{align}

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.

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.

• Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ ↵
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Systems of Linear Equations

## Solve Systems of Linear Equations with Two Variables

Learning objectives.

By the end of this section, you will be able to:

• Determine whether an ordered pair is a solution of a system of equations
• Solve a system of linear equations by graphing
• Solve a system of equations by substitution
• Solve a system of equations by elimination
• Choose the most convenient method to solve a system of linear equations

Before you get started, take this readiness quiz.

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .

When two or more linear equations are grouped together, they form a system of linear equations .

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

The steps to use to solve a system of linear equations by graphing are shown here.

• Graph the first equation.
• Graph the second equation on the same rectangular coordinate system.
• Determine whether the lines intersect, are parallel, or are the same line.

If the lines intersect, identify the point of intersection. This is the solution to the system.

If the lines are parallel, the system has no solution.

• If the lines are the same, the system has an infinite number of solutions.
• Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

no solution

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

infinitely many solutions

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.

Coincident lines have the same slope and same y- intercept.

The systems of equations in (Figure) and (Figure) each had two intersecting lines. Each system had one solution.

In (Figure) , the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like (Figure) , has no solution. We call a system of equations like this inconsistent. It has no solution.

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and (Figure) .

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ We will compare the slopes and intercepts of the two lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

• Solve one of the equations for either variable.
• Substitute the expression from Step 1 into the other equation.
• Solve the resulting equation.
• Substitute the solution in Step 3 into either of the original equations to find the other variable.
• Write the solution as an ordered pair.
• Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

We need to solve one equation for one variable. We will solve the first equation for y .

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d .

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

Then rewrite the system of equations.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

The steps are listed here for easy reference.

• Write both equations in standard form. If any coefficients are fractions, clear them.

Decide which variable you will eliminate.

• Multiply one or both equations so that the coefficients of that variable are opposites.
• Add the equations resulting from Step 2 to eliminate one variable.
• Solve for the remaining variable.
• Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

Since both equations are in standard form, using elimination will be most convenient.

Since one equation is already solved for y , using substitution will be most convenient.

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for x , using substitution will be most convenient.

ⓐ Since one equation is already solved for y , using substitution will be most convenient. ⓑ Since both equations are in standard form, using elimination will be most convenient.

## Practice Makes Perfect

In the following exercises, determine if the following points are solutions to the given system of equations.

In the following exercises, solve the following systems of equations by graphing.

infinite solutions

1 point, consistent and independent

infinite solutions, consistent, dependent

In the following exercises, solve the systems of equations by substitution.

In the following exercises, solve the systems of equations by elimination.

infinitely many

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ substitution ⓑ elimination

ⓐ elimination ⓑ substituion

## Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

ⓐ by graphing ⓑ by substitution

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Who can you ask for help? Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no – I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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