algebra 1 assignment solve each system by graphing

Solving Systems By Graphing

Popular tutorials in solving systems by graphing.

How Do You Solve a System of Equations by Graphing?

There are many different ways to solve a system of linear equations. In this tutorial, you'll see how to solve a system of linear equations by graphing both lines and finding their intersection. Take a look!

What's a System of Linear Equations?

A system of equations is a set of equations with the same variables. If the equations are all linear, then you have a system of linear equations! To solve a system of equations, you need to figure out the variable values that solve all the equations involved. This tutorial will introduce you to these systems.

What's a Solution to a System of Linear Equations?

If you have a system of equations that contains two equations with the same two unknown variables, then the solution to that system is the ordered pair that makes both equations true at the same time. Follow along as this tutorial uses an example to explain the solution to a system of equations!

What's a Consistent Independent System of Equations?

Looking at a system of equations with only one solution? That means that those equations intersect only at that one point. That kind of solution is called consistent and independent! This tutorial explains systems with one solution and even shows you an example!

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4.1: Solving Linear Systems by Graphing

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Learning Objectives

• Check solutions to systems of linear equations.
• Solve linear systems using the graphing method.
• Identify dependent and inconsistent systems.

Definition of a Linear System

Real-world applications are often modeled using more than one variable and more than one equation. A system of equations consists of a set of two or more equations with the same variables. In this section, we will study linear systems consisting of two linear equations each with two variables. For example,

A solution to a linear system, or simultaneous solution, to a linear system is an ordered pair \((x, y)\) that solves both of the equations. In this case, \((3, 2)\) is the only solution. To check that an ordered pair is a solution, substitute the corresponding \(x\)-and \(y\)-values into each equation and then simplify to see if you obtain a true statement for both equations.

\(\color{Cerulean}{Check:}\:\color{black}{(3,2)}\)

\(\begin{array} {c|c}{Equation\:1:\:\:2x-37=0}&{Equation\:2\:\:-4x+2y=-8}\\{2(\color{OliveGreen}{3}\color{black}{)-3(}\color{OliveGreen}{2}\color{black}{)=0}}&{-4(\color{OliveGreen}{3}\color{black}{)+2(}\color{OliveGreen}{2}\color{black}{)=-8}}\\{6-6=0}&{-12+4=-8}\\{0=0\quad\color{Cerulean}{\checkmark}}&{-8=-8\quad\color{Cerulean}{\checkmark}} \end{array}\)

Example \(\PageIndex{1}\)

Determine whether \((1, 0)\) is a solution to the system

\(\left\{\begin{aligned} x−y&=1 \\ −2x+3y&=5 \end{aligned}\right.\).

Substitute the appropriate values into both equations.

\(\color{Cerulean}{Check:}\:\color{black}{(1,0)}\)

\(\begin{array}{c|c} {Equation\:1:\:\:x-y=1}&{Equation\:2:\:\:-2x+3y=5}\\{(\color{OliveGreen}{1}\color{black}{)-(}\color{OliveGreen}{0}\color{black}{)=1}}&{-2(\color{OliveGreen}{1}\color{black}{)+3(}\color{OliveGreen}{0}\color{black}{)=5}}\\{1-0=1}&{-2+0=5}\\{1=1\quad\color{Cerulean}{\checkmark}}&{-2=5\quad\color{red}{x}} \end{array}\)

Since \((1, 0)\) does not satisfy both equations, it is not a solution.

Exercise \(\PageIndex{1}\)

Is \((−2, 4)\) a solution to the system

\(\begin{aligned} x−y&=−6\\−2x+3y&=16\end{aligned}\)?

Solve by Graphing

Geometrically, a linear system consists of two lines, where a solution is a point of intersection. To illustrate this, we will graph the following linear system with a solution of \((3, 2)\):

First, rewrite the equations in slope-intercept form so that we may easily graph them.

\(\begin{array}{c|c}{2x-3y=0}&{-4x+2y=-8}\\{2x-3y\color{Cerulean}{-2x}\color{black}{=0}\color{Cerulean}{-2x}}&{-4x+2y\color{Cerulean}{+4x}\color{black}{=-8}\color{Cerulean}{+4x}}\\{-3y=-2x}&{2y=4x-8}\\{\frac{-3y}{\color{Cerulean}{-3}}\color{black}{=\frac{-2x}{\color{Cerulean}{-3}}}}&{\frac{2y}{\color{Cerulean}{2}}\color{black}{=\frac{4x-8}{\color{Cerulean}{2}}}}\\{y=\frac{2}{3}x}&{y=2x-4} \end{array}\)

Next, replace these forms of the original equations in the system to obtain what is called an equivalent system. Equivalent systems share the same solution set.

\(\color{Cerulean}{Original\:system}\qquad\color{Cerulean}{Equivalent\:system}\)

If we graph both of the lines on the same set of axes, then we can see that the point of intersection is indeed \((3, 2)\), the solution to the system.

To summarize, linear systems described in this section consist of two linear equations each with two variables. A solution is an ordered pair that corresponds to a point where the two lines in the rectangular coordinate plane intersect. Therefore, we can solve linear systems by graphing both lines on the same set of axes and determining the point where they cross. When graphing the lines, take care to choose a good scale and use a straightedge to draw the line through the points; accuracy is very important here. The steps for solving linear systems using the graphing method are outlined in the following example.

Example \(\PageIndex{2}\)

Solve by graphing:

\(\left\{\begin{aligned} x−y&=−4 \\2x+y&=1\end{aligned}\right.\).

Step 1 : Rewrite the linear equations in slope-intercept form.

\(\begin{array}{c|c}{x-y=-4}&{2x+y=1}\\{x-y\color{Cerulean}{-x}\color{black}{=-4}\color{Cerulean}{-x}}&{2x+y\color{Cerulean}{-2x}\color{black}{=1}\color{Cerulean}{-2x}}\\{-y=-x-4}&{y=-2x+1}\\{\frac{-y}{\color{Cerulean}{-1}}\color{black}{=\frac{-x-4}{\color{Cerulean}{-1}}}}&{}\\{y=x+4}&{} \end{array}\)

Step 2 : Write the equivalent system and graph the lines on the same set of axes.

\(\begin{array}{c|c} {\color{Cerulean}{Line\:1:}\:\color{black}{y=x+4}}&{\color{Cerulean}{Line\:2}\:\color{black}{y=-2x+1}}\\{y-intercept:\:(0,4)}&{y-intercept:\:(0,1)}\\{slope:\:m=1=\frac{1}{1}=\frac{rise}{run}}&{slope:\:m=-2=\frac{-2}{1}=\frac{rise}{run}} \end{array}\)

Step 3 : Use the graph to estimate the point where the lines intersect and check to see if it solves the original system. In the above graph, the point of intersection appears to be \((−1, 3)\).

\(\color{Cerulean}{Check:}\:\color{black}{(-1,3)}\)

\(\begin{array}{c|c} {Line\:1:\:\: x-y=-4}&{Line\:2:\:\: 2x+y=1}\\{(\color{OliveGreen}{-1}\color{black}{)-(}\color{OliveGreen}{3}\color{black}{)=-4}}&{2(\color{OliveGreen}{-1}\color{black}{)+(}\color{OliveGreen}{3}\color{black}{)=1}}\\{-1-3=-4}&{-2+3=1}\\{-4=-4\quad\color{Cerulean}{\checkmark}}&{1=1\quad\color{Cerulean}{\checkmark}} \end{array}\)

Example \(\PageIndex{3}\)

\(\left\{\begin{aligned} 2x+y&=2\\−2x+3y&=−18 \end{aligned}\right.\).

We first solve each equation for \(y\) to obtain an equivalent system where the lines are in slope-intercept form.

Graph the lines and determine the point of intersection.

\(\color{Cerulean}{Check:}\:\:\color{black}{(3,-4)}\)

\(\begin{array} {c|c} {2x+y=2}&{-2x+3y=-18}\\{2(\color{OliveGreen}{3}\color{black}{)+(}\color{OliveGreen}{-4}\color{black}{)=2}}&{-2(\color{OliveGreen}{3}\color{black}{)+3(}\color{OliveGreen}{-4}\color{black}{)=-18}}\\{6-4=2}&{-6-12=-18}\\{2=2\quad\color{Cerulean}{\checkmark}}&{-18=-18\quad\color{Cerulean}{\checkmark}} \end{array}\)

Example \(\PageIndex{4}\)

\(\left\{\begin{aligned} 3x+y&=6\\y&=−3 \end{aligned}\right.\)

\(\color{Cerulean}{Check:}\:\color{black}{(3,-3)}\)

\(\begin{array}{c|c}{3x+y=6}&{y=-3}\\{3(\color{OliveGreen}{3}\color{black}{)+(}\color{OliveGreen}{-3}\color{black}{)=6}}&{(\color{OliveGreen}{-3}\color{black}{)=-3}}\\{9-3=6}&{-3=-3\quad\color{Cerulean}{\checkmark}}\\{6=6\quad\color{Cerulean}{\checkmark}}&{} \end{array}\)

The graphing method for solving linear systems is not ideal when the solution consists of coordinates that are not integers. There will be more accurate algebraic methods in sections to come, but for now, the goal is to understand the geometry involved when solving systems. It is important to remember that the solutions to a system correspond to the point, or points, where the graphs of the equations intersect.

Exercise \(\PageIndex{2}\)

\(\left\{\begin{aligned}−x+y&=6\\ 5x+2y&=−2 \end{aligned}\right.\).

Dependent and Inconsistent Systems

Systems with at least one solution are called consistent systems. Up to this point, all of the examples have been of consistent systems with exactly one ordered pair solution. It turns out that this is not always the case. Sometimes systems consist of two linear equations that are equivalent. If this is the case, the two lines are the same and when graphed will coincide. Hence the solution set consists of all the points on the line. This is a dependent system. Given a consistent linear system with two variables, there are two possible results:

The solutions to independent systems are ordered pairs \((x, y)\). We need some way to express the solution sets to dependent systems, since these systems have infinitely many solutions, or points of intersection. Recall that any line can be written in slope-intercept form, \(y=mx+b\). Here, \(y\) depends on \(x\). So we may express all the ordered pair solutions \((x, y)\) in the form \((x, mx+b)\), where \(x\) is any real number.

Example \(\PageIndex{5}\)

\(\left\{\begin{aligned}−2x+3y&=−9 \\ 4x−6y&=18\end{aligned}\right.\).

Determine slope-intercept form for each linear equation in the system.

\(\begin{array}{c|c} {-2x+3y=-9}&{4x-6y=18}\\{-2x+3y\color{Cerulean}{+2x}\color{black}{=-9}\color{Cerulean}{+2x}}&{4x-6y\color{Cerulean}{-4x}\color{black}{=18}\color{Cerulean}{-4x}}\\{3y=2x-9}&{-6y=-4x+18}\\{\frac{3y}{\color{Cerulean}{3}}\color{black}{=\frac{2x-9}{\color{Cerulean}{3}}}}&{\frac{-6y}{\color{Cerulean}{-6}}\color{black}{=\frac{-4x+18}{\color{Cerulean}{-6}}}}\\{y=\frac{2}{3}x-3}&{y=\frac{-4}{-6}x+\frac{18}{-6}}\\{}&{y=\frac{2}{3}x-3} \end{array}\)

In slope-intercept form, we can easily see that the system consists of two lines with the same slope and same \(y\)-intercept.

They are, in fact, the same line. And the system is dependent.

\((x,\frac{2}{3}x-3)\)

In this example, it is important to notice that the two lines have the same slope and same \(y\)-intercept. This tells us that the two equations are equivalent and that the simultaneous solutions are all the points on the line \(y=\frac{2}{3}x−3\). This is a dependent system, and the infinitely many solutions are expressed using the form \((x, mx+b)\). Other resources may express this set using set notation, \(\{(x, y) | y=\frac{2}{3}x−3\}\), which reads “the set of all ordered pairs \((x, y)\) such that \(y\) equals two-thirds \(x\) minus \(3\).” Sometimes the lines do not cross and there is no point of intersection. Such systems have no solution, \(Ø\), and are called inconsistent systems.

Example \(\PageIndex{6}\)

\(\left\{\begin{aligned}−2x+5y&=−15\\−4x+10y&=10\end{aligned}\right.\).

Determine slope-intercept form for each linear equation.

\(\begin{array}{c|c} {-2x+5y=-15}&{-4x+10y=10}\\{-2x+5y\color{Cerulean}{+2x}\color{black}{=-15}\color{Cerulean}{+2x}}&{-4x+10y\color{Cerulean}{+4x}\color{black}{=10}\color{Cerulean}{+4x}}\\{5y=2x-15}&{10y=4x+10}\\{\frac{5y}{\color{Cerulean}{5}}\color{black}{=\frac{2x-15}{\color{Cerulean}{5}}}}&{\frac{10y}{\color{Cerulean}{10}}\color{black}{=\frac{4x+10}{\color{Cerulean}{10}}}}\\{y=\frac{2}{5}x-3}&{y=\frac{2}{5}x+1} \end{array}\)

In slope-intercept form, we can easily see that the system consists of two lines with the same slope and different \(y\)-intercepts.

Therefore, they are parallel and will never intersect.

There is no simultaneous solution, \(Ø\).

Exercise \(\PageIndex{3}\)

\(\left\{\begin{aligned} x+y=&−1\\−2x−2y&=2 \end{aligned}\right.\).

\((x,-x-1)\)

Key Takeaways

• In this section, we limit our study to systems of two linear equations with two variables. Solutions to such systems, if they exist, consist of ordered pairs that satisfy both equations. Geometrically, solutions are the points where the graphs intersect.
• The graphing method for solving linear systems requires us to graph both of the lines on the same set of axes as a means to determine where they intersect.
• The graphing method is not the most accurate method for determining solutions, particularly when the solutions have coordinates that are not integers. It is a good practice to always check your solutions.
• Some linear systems have no simultaneous solution. These systems consist of equations that represent parallel lines with different \(y\)-intercepts and do not intersect in the plane. They are called inconsistent systems and the solution set is the empty set, \(Ø\).
• Some linear systems have infinitely many simultaneous solutions. These systems consist of equations that are equivalent and represent the same line. They are called dependent systems and their solutions are expressed using the notation \((x, mx+b)\), where \(x\) is any real number.

Exercise \(\PageIndex{4}\) Solutions to Linear Systems

Determine whether the given ordered pair is a solution to the given system.

• \((3, −2); \left\{\begin{aligned} x+y&=-1\\-2x-2y&=2 \end{aligned}\right.\)
• \((−5, 0); \left\{\begin{aligned} x+y&=−1\\−2x−2y&=2 \end{aligned}\right.\)
• \((−2, −6); \left\{\begin{aligned}−x+y&=−4\\3x−y&=−12 \end{aligned}\right.\)
• \((2, −7); \left\{\begin{aligned} 3x+2y&=−8\\−5x−3y&=11 \end{aligned}\right.\)
• \((0, −3); \left\{\begin{aligned}5x−5y&=15\\−13x+2y&=−6 \end{aligned}\right.\)
• \((−12, 14); \left\{\begin{aligned} x+y&=−14\\−2x−4y&=0 \end{aligned}\right.\)
• \((\frac{3}{4}, \frac{1}{4}); \left\{\begin{aligned} −x−y&=−1\\−4x−8y&=5 \end{aligned}\right.\)
• \((−3, 4); \left\{\begin{aligned} \frac{1}{3}x+\frac{1}{2}y&=1 \\ \frac{2}{3}x−\frac{3}{2}y&=−8 \end{aligned}\right.\)
• \((−5, −3); \left\{\begin{aligned} y&=−35x−10 \\ y&=5 \end{aligned}\right.\)
• \((4, 2); \left\{\begin{aligned} x&=4−7\\x+4y&=8 \end{aligned}\right.\)

Exercise \(\PageIndex{5}\) Solutions to Linear Systems

Given the graph, determine the simultaneous solution.

1. \((5, 0)\)

3. \((2, 1)\)

5. \((0, 0)\)

7. \((x, 2x−2)\)

9. \(∅\)

Exercise \(\PageIndex{6}\) Solving Linear Systems

Solve by graphing.

• \(\left\{\begin{aligned} y &=\frac{3}{2}x + 6\\y&=−x + 1 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y& =\frac{3}{4}x + 2\\y&=−\frac{1}{4}x − 2 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y& =x − 4\\y&=−x + 2 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y&=− 5 x + 4\\y& = 4 x − 5 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y& =\frac{2}{5} x + 1\\ y& =\frac{3}{5} x \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=−\frac{2}{5} x + 6\\y& =\frac{2}{5} x +10 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y&=− 2\\y& =x + 1 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y& = 3\\ x&=− 3 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y& = 0\\y& =\frac{2}{5} x − 4 \end{aligned}\right.\)
• \(\left\{\begin{aligned} x& = 2 \\y& = 3 x \end{aligned}\right.\)
• \(\left\{\begin{aligned} y& =\frac{3}{5} x − 6\\y& =\frac{3}{5} x − 3 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y&=−\frac{1}{2}x + 1\\ y&=−\frac{1}{2}x + 1 \end{aligned}\right.\)
• \(\left\{\begin{aligned}2 x + 3 y &=18 \\ − 6 x + 3 y&=− 6 \end{aligned}\right.\)
• \(\left\{\begin{aligned} − 3 x + 4y &=20\\2 x + 8y &= 8\end{aligned}\right.\)
• \(\left\{\begin{aligned} − 2 x +y &=1 \\2 x − 3 y& = 9\end{aligned}\right.\)
• \(\left\{\begin{aligned} x + 2y&=−8\\5 x + 4y&=− 4\end{aligned}\right.\)
• \(\left\{\begin{aligned}4 x + 6y &=36\\2 x − 3 y &= 6\end{aligned}\right.\)
• \(\left\{\begin{aligned} 2 x − 3 y &=18\\6 x − 3 y&=− 6\end{aligned}\right.\)
• \(\left\{\begin{aligned} 3 x + 5y &=30\\ − 6 x −10y&=−10\end{aligned}\right.\)
• \(\left\{\begin{aligned}−x + 3 y &=3\\5 x −15 y&=−15\end{aligned}\right.\)
• \(\left\{\begin{aligned}x −y &= 0\\ −x +y &= 0\end{aligned}\right.\)
• \(\left\{\begin{aligned}y &=x\\y −x& = 1\end{aligned}\right.\)
• \(\left\{\begin{aligned}3 x + 2y &= 0\\ x &= 2\end{aligned}\right.\)
• \(\left\{\begin{aligned}2 x +\frac{1}{3}y &=\frac{2}{3}\\ − 3 x +12y&=− 2\end{aligned}\right.\)
• \(\left\{\begin{aligned}\frac{1}{10}x +\frac{1}{5}y &= 2\\ −\frac{1}{5} x +\frac{1}{5}y&=− 1\end{aligned}\right.\)
• \(\left\{\begin{aligned}\frac{1}{3} x −\frac{1}{2}y &= 1 \\ \frac{1}{3} x +\frac{1}{5}y& = 1 \end{aligned}\right.\)
• \(\left\{\begin{aligned} \frac{1}{9}x +\frac{1}{6}y &= 0 \\ \frac{1}{9}x +\frac{1}{4}y &=\frac{1}{2}\end{aligned}\right.\)
• \(\left\{\begin{aligned} \frac{5}{16}x −\frac{1}{2}y &= 5\\ −\frac{5}{16}x +\frac{1}{2}y &=\frac{5}{2} \end{aligned}\right.\)
• \(\left\{\begin{aligned} \frac{1}{6}x−\frac{1}{2}y&=\frac{9}{2} \\ −\frac{1}{18}x+\frac{1}{6}y&=−\frac{3}{2} \end{aligned}\right.\)
• \(\left\{\begin{aligned} \frac{1}{2}x−\frac{1}{4}y&=−\frac{1}{2} \\ \frac{1}{3}x−\frac{1}{2}y&=3\end{aligned}\right.\)
• \(\left\{\begin{aligned} y&=4\\x&=−5 \end{aligned}\right.\)
• \(\left\{\begin{aligned} y&=−3\\x&=2\end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=0\\x&=0\end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=−2\\y&=3\end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=5\\y&=−5\end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=2\\y−2&=0\end{aligned}\right.\)
• \(\left\{\begin{aligned}x&=−5\\x&=1\end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=x\\x&=0\end{aligned}\right.\)
• \(\left\{\begin{aligned}4x+6y&=3\\−x+y&=−2\end{aligned}\right.\)
• \(\left\{\begin{aligned}−2x+20y&=20\\3x+10y&=−10\end{aligned}\right.\)

1. \((−2, 3)\)

3. \((3, −1)\)

5. \((5, 3)\)

7. \((−3, −2)\)

9. \((10, 0)\)

11. \(∅\)

13. \((3, 4)\)

15. \((−3, −5)\)

17. \((6, 2)\)

19. \(∅\)

21. \((x, x)\)

23. \((2, −3)\)

25. \((10, 5)\)

27. \((−9, 6)\)

29. \((x, 13x−9)\)

31. \((−5, 4)\)

33. \((0, 0)\)

35. \(∅\)

37. \(∅\)

39. \((\frac{3}{2}, −\frac{1}{2})\)

Exercise \(\PageIndex{7}\) Solving Linear Systems

Set up a linear system of two equations and two variables and solve it using the graphing method.

• The sum of two numbers is
• The larger number is \(10\) less than five times the smaller.
• The difference between two numbers is \(12\) and their sum is \(4\).
• Where on the graph of \(3x−2y=6\) does the \(x\)-coordinate equal the \(y\)-coordinate?
• Where on the graph of \(−5x+2y=30\) does the \(x\)-coordinate equal the \(y\)-coordinate?

1. The two numbers are \(5\) and \(15\).

3. \((6, 6)\)

Exercise \(\PageIndex{8}\) Solving Linear Systems

A regional bottled water company produces and sells bottled water. The following graph depicts the supply and demand curves of bottled water in the region. The horizontal axis represents the weekly tonnage of product produced, \(Q\). The vertical axis represents the price per bottle in dollars, \(P\).

Use the graph to answer the following questions.

• Determine the price at which the quantity demanded is equal to the quantity supplied.
• If production of bottled water slips to \(20\) tons, then what price does the demand curve predict for a bottle of water?
• If production of bottled water increases to \(40\) tons, then what price does the demand curve predict for a bottle of water?
• If the price of bottled water is set at $\(2.50\) dollars per bottle, what quantity does the demand curve predict?

1. $\(1.25\)

3. $\(1.00\)

Exercise \(\PageIndex{9}\) Discussion Board Topics

• Discuss the weaknesses of the graphing method for solving systems.
• Explain why the solution set to a dependent linear system is denoted by \((x, mx+ b)\).

1. Answers may vary

5.1 Solve Systems of Equations by Graphing

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
• Determine the number of solutions of linear system
• Solve applications of systems of equations by graphing

Be Prepared 5.1

Before you get started, take this readiness quiz.

For the equation y = 2 3 x − 4 y = 2 3 x − 4 ⓐ is ( 6 , 0 ) ( 6 , 0 ) a solution? ⓑ is ( −3 , −2 ) ( −3 , −2 ) a solution? If you missed this problem, review Example 2.1 .

Be Prepared 5.2

Find the slope and y -intercept of the line 3 x − y = 12 3 x − y = 12 . If you missed this problem, review Example 4.42 .

Be Prepared 5.3

Find the x - and y -intercepts of the line 2 x − 3 y = 12 2 x − 3 y = 12 . If you missed this problem, review Example 4.21 .

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

In Solving Linear Equations and Inequalities we learned how to solve linear equations with one variable. Remember that the solution of an equation is a value of the variable that makes a true statement when substituted into the equation.

Now we will work with systems of linear equations , two or more linear equations grouped together.

System of Linear Equations

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

We will focus our work here on systems of two linear equations in two unknowns. Later, you may solve larger systems of equations.

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.

A linear equation in two variables, like 2 x + y = 7, has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) that make both equations true. These are called the solutions to a system of equations .

Solutions of a System of Equations

Solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ).

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.

Let’s consider the system below:

Is the ordered pair ( 2 , −1 ) ( 2 , −1 ) a solution?

The ordered pair (2, −1) made both equations true. Therefore (2, −1) is a solution to this system.

Let’s try another ordered pair. Is the ordered pair (3, 2) a solution?

The ordered pair (3, 2) made one equation true, but it made the other equation false. Since it is not a solution to both equations, it is not a solution to this system.

Example 5.1

Determine whether the ordered pair is a solution to the system: { x − y = −1 2 x − y = −5 { x − y = −1 2 x − y = −5

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

Determine whether the ordered pair is a solution to the system: { 3 x + y = 0 x + 2 y = −5 . { 3 x + y = 0 x + 2 y = −5 .

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

Determine whether the ordered pair is a solution to the system: { x − 3 y = −8 −3 x − y = 4 . { x − 3 y = −8 −3 x − y = 4 .

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

In this chapter 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 in Figure 5.2 :

For the first example of solving a system of linear equations in this section and in the next two sections, we will solve the same system of two linear equations. But we’ll use a different method in each section. After seeing the third method, you’ll decide which method was the most convenient way to solve this system.

Example 5.2

How to solve a system of linear equations by graphing.

Solve the system by graphing: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Solve the system by graphing: { x − 3 y = −3 x + y = 5 . { x − 3 y = −3 x + y = 5 .

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

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

To solve a system of linear equations by graphing.

• Step 1. Graph the first equation.
• Step 2. Graph the second equation on the same rectangular coordinate system.
• Step 3. Determine whether the lines intersect, are parallel, or are the same line.
• If the lines intersect, identify the point of intersection. Check to make sure it is a solution to both equations. 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.

Example 5.3

Solve the system by graphing: { y = 2 x + 1 y = 4 x − 1 . { y = 2 x + 1 y = 4 x − 1 .

Both of the equations in this system are in slope-intercept form, so we will use their slopes and y -intercepts to graph them. { y = 2 x + 1 y = 4 x − 1 { y = 2 x + 1 y = 4 x − 1

Solve each system by graphing: { y = 2 x + 2 y = − x − 4 . { y = 2 x + 2 y = − x − 4 .

Solve each system by graphing: { y = 3 x + 3 y = − x + 7 . { y = 3 x + 3 y = − x + 7 .

Both equations in Example 5.3 were given in slope–intercept form. This made it easy for us to quickly graph the lines. In the next example, we’ll first re-write the equations into slope–intercept form.

Example 5.4

Solve the system by graphing: { 3 x + y = −1 2 x + y = 0 . { 3 x + y = −1 2 x + y = 0 .

We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y -intercepts. { 3 x + y = −1 2 x + y = 0 { 3 x + y = −1 2 x + y = 0

Solve each system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve each system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

Usually when equations are given in standard form, the most convenient way to graph them is by using the intercepts. We’ll do this in Example 5.5 .

Example 5.5

Solve the system by graphing: { x + y = 2 x − y = 4 . { x + y = 2 x − y = 4 .

We will find the x - and y -intercepts of both equations and use them to graph the lines.

Solve each system by graphing: { x + y = 6 x − y = 2 . { x + y = 6 x − y = 2 .

Try It 5.10

Solve each system by graphing: { x + y = 2 x − y = −8 . { x + y = 2 x − y = −8 .

Do you remember how to graph a linear equation with just one variable? It will be either a vertical or a horizontal line.

Example 5.6

Solve the system by graphing: { y = 6 2 x + 3 y = 12 . { y = 6 2 x + 3 y = 12 .

Try It 5.11

Solve each system by graphing: { y = −1 x + 3 y = 6 . { y = −1 x + 3 y = 6 .

Try It 5.12

Solve each system by graphing: { x = 4 3 x − 2 y = 24 . { x = 4 3 x − 2 y = 24 .

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.

Example 5.7

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Try It 5.13

Solve each system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

Try It 5.14

Solve each system by graphing: { y = 3 x − 1 6 x − 2 y = 6 . { y = 3 x − 1 6 x − 2 y = 6 .

Example 5.8

Solve the system by graphing: { y = 2 x − 3 −6 x + 3 y = − 9 . { y = 2 x − 3 −6 x + 3 y = − 9 .

Try It 5.15

Solve each system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

Try It 5.16

Solve each system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

If you write the second equation in Example 5.8 in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

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

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

Determine the Number of Solutions of a Linear System

There will be times when we will want to know how many solutions there will be to a system of linear equations, but we might not actually have to find the solution. It will be helpful to determine this without graphing.

We have seen that two lines in the same plane must either intersect or are parallel. The systems of equations in Example 5.2 through Example 5.6 all had two intersecting lines. Each system had one solution.

A system with parallel lines, like Example 5.7 , has no solution. What happened in Example 5.8 ? The equations have coincident lines , and so the system had infinitely many solutions.

We’ll organize these results in Figure 5.3 below:

Parallel lines have the same slope but different y -intercepts. So, if we write both equations in a system of linear equations in slope–intercept form, we can see how many solutions there will be without graphing! Look at the system we solved in Example 5.7 .

The two lines have the same slope but different y -intercepts. They are parallel lines.

Figure 5.4 shows how to determine the number of solutions of a linear system by looking at the slopes and intercepts.

Let’s take one more look at our equations in Example 5.7 that gave us parallel lines.

When both lines were in slope-intercept form we had:

Do you recognize that it is impossible to have a single ordered pair ( x , y ) ( x , y ) that is a solution to both of those equations?

We call a system of equations like this an inconsistent system . It has no solution.

A system of equations that has at least one solution is called a consistent system .

Consistent and Inconsistent Systems

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 equations , 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.

Independent and Dependent Equations

Two equations are independent if they have different solutions.

Two equations are dependent if all the solutions of one equation are also solutions of the other equation.

Let’s sum this up by looking at the graphs of the three types of systems. See Figure 5.5 and Figure 5.6 .

Example 5.9

Without graphing, determine the number of solutions and then classify the system of equations: { y = 3 x − 1 6 x − 2 y = 12 . { y = 3 x − 1 6 x − 2 y = 12 .

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

Try It 5.17

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

{ y = − 2 x − 4 4 x + 2 y = 9 { y = − 2 x − 4 4 x + 2 y = 9

Try It 5.18

{ y = 1 3 x − 5 x − 3 y = 6 { y = 1 3 x − 5 x − 3 y = 6

Example 5.10

Without graphing, determine the number of solutions and then classify the system of equations: { 2 x + y = − 3 x − 5 y = 5 . { 2 x + y = − 3 x − 5 y = 5 .

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

Try It 5.19

{ 3 x + 2 y = 2 2 x + y = 1 { 3 x + 2 y = 2 2 x + y = 1

Try It 5.20

{ x + 4 y = 12 − x + y = 3 { x + 4 y = 12 − x + y = 3

Example 5.11

Without graphing, determine the number of solutions and then classify the system of equations. { 3 x − 2 y = 4 y = 3 2 x − 2 { 3 x − 2 y = 4 y = 3 2 x − 2

A system of equations whose graphs are coincident lines has infinitely many solutions and is consistent and dependent.

Try It 5.21

{ 4 x − 5 y = 20 y = 4 5 x − 4 { 4 x − 5 y = 20 y = 4 5 x − 4

Try It 5.22

{ −2 x − 4 y = 8 y = − 1 2 x − 2 { −2 x − 4 y = 8 y = − 1 2 x − 2

Solve Applications of Systems of Equations by Graphing

We will use the same problem solving strategy we used in Math Models to set up and solve applications of systems of linear equations. We’ll modify the strategy slightly here to make it appropriate for systems of equations.

Use a problem solving strategy for systems of linear equations.

• Step 1. Read the problem. Make sure all the words and ideas are understood.

Step 2. Identify what we are looking for.

Step 3. Name what we are looking for. Choose variables to represent those quantities.

Step 4. Translate into a system of equations.

Step 5. Solve the system of equations using good algebra techniques.

Step 6. Check the answer in the problem and make sure it makes sense.

Step 7. Answer the question with a complete sentence.

Step 5 is where we will use the method introduced in this section. We will graph the equations and find the solution.

Example 5.12

Sondra is making 10 quarts of punch from fruit juice and club soda. The number of quarts of fruit juice is 4 times the number of quarts of club soda. How many quarts of fruit juice and how many quarts of club soda does Sondra need?

Step 1. Read the problem.

We are looking for the number of quarts of fruit juice and the number of quarts of club soda that Sondra will need.

  Let f = f = number of quarts of fruit juice.      c = c = number of quarts of club soda

We now have the system. { f + c = 10 f = 4 c { f + c = 10 f = 4 c

The point of intersection (2, 8) is the solution. This means Sondra needs 2 quarts of club soda and 8 quarts of fruit juice.

Does this make sense in the problem?

Yes, the number of quarts of fruit juice, 8 is 4 times the number of quarts of club soda, 2.

Yes, 10 quarts of punch is 8 quarts of fruit juice plus 2 quarts of club soda.

Sondra needs 8 quarts of fruit juice and 2 quarts of soda.

Try It 5.23

Manny is making 12 quarts of orange juice from concentrate and water. The number of quarts of water is 3 times the number of quarts of concentrate. How many quarts of concentrate and how many quarts of water does Manny need?

Try It 5.24

Alisha is making an 18 ounce coffee beverage that is made from brewed coffee and milk. The number of ounces of brewed coffee is 5 times greater than the number of ounces of milk. How many ounces of coffee and how many ounces of milk does Alisha need?

Access these online resources for additional instruction and practice with solving systems of equations by graphing.

• Instructional Video Solving Linear Systems by Graphing
• Instructional Video Solve by Graphing

Section 5.1 Exercises

Practice makes perfect.

Determine Whether an Ordered Pair is a Solution of a System of Equations . In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ 7 x − 4 y = −1 −3 x − 2 y = 1 { 7 x − 4 y = −1 −3 x − 2 y = 1

ⓐ ( 1 , 2 ) ( 1 , 2 ) ⓑ ( 1 , −2 ) ( 1 , −2 )

{ 2 x + y = 5 x + y = 1 { 2 x + y = 5 x + y = 1

ⓐ ( 4 , −3 ) ( 4 , −3 ) ⓑ ( 2 , 0 ) ( 2 , 0 )

{ −3 x + y = 8 − x + 2 y = −9 { −3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ x + y = 1 y = 2 5 x { x + y = 1 y = 2 5 x

ⓐ ( 5 7 , 2 7 ) ( 5 7 , 2 7 ) ⓑ ( 5 , 2 ) ( 5 , 2 )

{ x + 5 y = 10 y = 3 5 x + 1 { x + 5 y = 10 y = 3 5 x + 1

ⓐ ( −10 , 4 ) ( −10 , 4 ) ⓑ ( 5 4 , 7 4 ) ( 5 4 , 7 4 )

{ x + 3 y = 9 y = 2 3 x − 2 { x + 3 y = 9 y = 2 3 x − 2

ⓐ ( −6 , 5 ) ( −6 , 5 ) ⓑ ( 5 , 4 3 ) ( 5 , 4 3 )

Solve a System of Linear Equations by Graphing In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ −3 x + y = −1 2 x + y = 4 { −3 x + y = −1 2 x + y = 4

{ −2 x + 3 y = −3 x + y = 4 { −2 x + 3 y = −3 x + y = 4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ − x + y = −3 4 x + 4 y = 4 { − x + y = −3 4 x + 4 y = 4

{ x − y = 3 2 x − y = 4 { x − y = 3 2 x − y = 4

{ −3 x + y = −2 4 x − 2 y = 6 { −3 x + y = −2 4 x − 2 y = 6

{ x + y = 5 2 x − y = 4 { x + y = 5 2 x − y = 4

{ x − y = 2 2 x − y = 6 { x − y = 2 2 x − y = 6

{ x + y = 2 x − y = 0 { x + y = 2 x − y = 0

{ x + y = 6 x − y = −8 { x + y = 6 x − y = −8

{ x + y = −5 x − y = 3 { x + y = −5 x − y = 3

{ x + y = 4 x − y = 0 { x + y = 4 x − y = 0

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ −2 x + 3 y = 3 x + 3 y = 12 { −2 x + 3 y = 3 x + 3 y = 12

{ 2 x − y = 4 2 x + 3 y = 12 { 2 x − y = 4 2 x + 3 y = 12

{ 2 x + 3 y = 6 y = −2 { 2 x + 3 y = 6 y = −2

{ −2 x + y = 2 y = 4 { −2 x + y = 2 y = 4

{ x − 3 y = −3 y = 2 { x − 3 y = −3 y = 2

{ 2 x − 2 y = 8 y = −3 { 2 x − 2 y = 8 y = −3

{ 2 x − y = −1 x = 1 { 2 x − y = −1 x = 1

{ x + 2 y = 2 x = −2 { x + 2 y = 2 x = −2

{ x − 3 y = −6 x = −3 { x − 3 y = −6 x = −3

{ x + y = 4 x = 1 { x + y = 4 x = 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 −2 x − 6 y = 3 { x + 3 y = 4 −2 x − 6 y = 3

{ −2 x + 4 y = 4 y = 1 2 x { −2 x + 4 y = 4 y = 1 2 x

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 −8 x − 4 y = −24 { 2 x + y = 6 −8 x − 4 y = −24

{ 5 x + 2 y = 7 −10 x − 4 y = −14 { 5 x + 2 y = 7 −10 x − 4 y = −14

{ x + 3 y = −6 4 y = − 4 3 x − 8 { x + 3 y = −6 4 y = − 4 3 x − 8

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

{ −3 x + 2 y = −2 y = − x + 4 { −3 x + 2 y = −2 y = − x + 4

{ − x + 2 y = −2 y = − x − 1 { − x + 2 y = −2 y = − x − 1

Determine the Number of Solutions of a Linear System Without graphing the following systems of equations, determine the number of solutions and then classify the system of equations.

{ y = 2 3 x + 1 −2 x + 3 y = 5 { y = 2 3 x + 1 −2 x + 3 y = 5

{ y = 1 3 x + 2 x − 3 y = 9 { y = 1 3 x + 2 x − 3 y = 9

{ y = −2 x + 1 4 x + 2 y = 8 { y = −2 x + 1 4 x + 2 y = 8

{ y = 3 x + 4 9 x − 3 y = 18 { y = 3 x + 4 9 x − 3 y = 18

{ y = 2 3 x + 1 2 x − 3 y = 7 { y = 2 3 x + 1 2 x − 3 y = 7

{ 3 x + 4 y = 12 y = −3 x − 1 { 3 x + 4 y = 12 y = −3 x − 1

{ 4 x + 2 y = 10 4 x − 2 y = −6 { 4 x + 2 y = 10 4 x − 2 y = −6

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ y = x + 1 − x + y = 1 { y = x + 1 − x + y = 1

{ y = 2 x + 3 2 x − y = −3 { y = 2 x + 3 2 x − y = −3

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

Solve Applications of Systems of Equations by Graphing In the following exercises, solve.

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Jamal is making a snack mix that contains only pretzels and nuts. For every ounce of nuts, he will use 2 ounces of pretzels. How many ounces of pretzels and how many ounces of nuts does he need to make 45 ounces of snack mix?

Enrique is making a party mix that contains raisins and nuts. For each ounce of nuts, he uses twice the amount of raisins. How many ounces of nuts and how many ounces of raisins does he need to make 24 ounces of party mix?

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

Everyday Math

Leo is planning his spring flower garden. He wants to plant tulip and daffodil bulbs. He will plant 6 times as many daffodil bulbs as tulip bulbs. If he wants to plant 350 bulbs, how many tulip bulbs and how many daffodil bulbs should he plant?

A marketing company surveys 1,200 people. They surveyed twice as many females as males. How many males and females did they survey?

Writing Exercises

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

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

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. Whom 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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Solve Linear Systems Algebra 1 Practice Packet

Also included in

Description

This packet contains 14 organized problems to practice solving linear systems using the following methods:

• Graphing (#1-4)
• Substitution (#5-8)
• Elimination (#9-14)

Note: No Solution and Infinitely Many Solutions both come up when solving by graphing and once for each algebraic method. All answers are integer coordinates.

An answer key is included!

This resource is also included in the following bundles:

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