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Systems of Linear Equations in Two Variables

Determine and show whether an ordered pair is a solution to a system of equations.

Learning Objectives

After completing this section, you should be able to:

Determine and show whether an ordered pair is a solution to a system of equations.
Solve systems of linear equations using graphical methods.
Solve systems of linear equations using substitution.
Solve systems of linear equations using elimination.
Identify systems with no solution or infinitely many solutions.
Solve applications of systems of linear equations.

Determining If an Ordered Pair Is a Solution to a System of Equations

When we solved linear equations in Linear Equations in One Variable with Applications and Linear Inequalities in One Variable with Applications, 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.

In this section, we will focus our work on systems of two linear equations in two unknowns (variables) and applications of systems of linear 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.

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

A linear equation in two variables, such as $2x+y=7$, 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 of 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.

Determining Whether an Ordered Pair Is a Solution to the System

Try it.

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

\[\{\begin{array}{lll}x-y & = & -1 \\ 2x-y & = & -5\end{array}\]

$(-2,-1)$
$(-4,-3)$

Solution

\[\{\begin{array}{lll}x-y & = & -1 \\ 2x-y & = & -5\end{array}\] We substitute $x=-2$ and $y=-1$ into both equations.
\[\begin{array}{l}x-y=-1\ \\ -2-(-1)\overset{?}{=}-1\ \\ -1=-1✓\end{array}\ \begin{array}{l}2x-y=-5 \\ 2(-2)-(-1)\overset{?}{=}-5 \\ -3\ne -5\end{array}\]
$(-2,-1)$ does not make both equations true. $(-2,-1)$ is not a solution.
We substitute $x=-4$ and $y=-3$ into both equations.
\[\begin{array}{l}x-y=-1\ \\ -4-(-3)\overset{?}{=}-1\ \\ -1=-1✓\end{array}\ \begin{array}{l}2x-y=-5\ \\ 2•(-4)-(-3)\overset{?}{=}-5\ \\ -5=-5✓\end{array}\]
$(-4,-3)$ makes both equations true. $(-4,-3)$ is a solution.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Systems of Linear Equations Using Graphical Methods

We will use three methods to solve a system of linear equations. The first method we will 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 points the lines have in common, we will find the solution to the system.

Most linear equations in one variable have one solution; but for some equations called contradictions, there are 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 .

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end you will 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.

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.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Systems of Linear Equations Using Substitution

We will now solve systems of linear equations by the substitution method. We will use the same system we used for graphing.

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

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. This process is summarized here:

Step 1: Solve one of the equations for either variable.

Step 2: Substitute the expression from Step 1 into the other equation.

Step 3: Solve the resulting equation.

Step 4: Substitute the solution in Step 3 into either of the original equations to find the other variable.

Solving a System of Linear Equations Using Substitution

Try it.

Solve this system of linear equations by substitution:

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

Solution

Step 1: Solve one of the equations for either variable.	We’ll solve the first equation for $y$.	$\begin{array}{lllll}\begin{array}{l}\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\end{array} \\ \begin{array}{lll}2x+y & = & 7 \\ y & = & 7-2x\end{array}\end{array}$
Step 2: Substitute the expression from Step 1 into the other equation.	We replace $y$ in the second equation with the expression $7-2x$.	$\begin{array}{l}x-2y=6 \\ x-2(7-2x)=6\end{array}$
Step 3: Solve the resulting equation.	Now we have an equation with just 1 variable. We know how to solve this!	$\begin{array}{lll}x-2(7-2x) & = & 6 \\ x-14+4x & = & 6 \\ 5x & = & 20 \\ x & = & 4\end{array}$
Step 4: Substitute the solution from Step 3 into one of the original equations to find the other variable.	We’ll use the first equation and replace $x$ with 4.	$\begin{array}{lll}2x+y & = & 7 \\ 2(4)+y & = & 7 \\ 8+y & = & 7 \\ y & = & -1\end{array}$
Step 5: Write the solution as an ordered pair.	The ordered pair is ($x$, $y$).	$(4,-1)$
Step 6: Check that the ordered pair is a solution to both original equations.	Substitute $x=4$, $y=-1$ into both equations and make sure they are both true.	$\begin{array}{llllllllllllll}\begin{array}{lll}2x+y & = & 7 \\ 2(4)+(-1) & \overset{?}{=} & 7 \\ 7 & = & 7✓\end{array} & \begin{array}{lll}x-2y & = & 6 \\ 4-2(-1) & \overset{?}{=} & 6 \\ 6 & = & 6✓\end{array}\end{array}$ Both equations are true. $(4,-1)$ is the solution to the system.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Systems of Linear Equations Using 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 $d$:

if $a=b$

and $c=d$

then $a+c=b+d$.

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:

\[\begin{array}{l}\{\begin{array}{l}3x+y=5 \\ \underset{_}{2x-y=0}\end{array} \\ 5x\ =5\ \end{array}\]\[\{\begin{array}{l}x+4y=-2 \\ 2x+5y=-2\end{array}\]\[\{\begin{array}{lll}-2(x+4y) & = & -2(-2) \\ 2x+5y & = & -2\end{array}\]\[\{\begin{array}{lll}-2x-8y & = & 4 \\ 2x+5y & = & -2\end{array}\]\[\begin{array}{l}\{\begin{array}{l}-2x-8y=4 \\ \underset{_}{2x+5y=-2\ }\end{array} \\ -3y=2\ \end{array}\]

Condensed — the full section is in OpenStax Contemporary Mathematics.

Identifying Systems with No Solution or Infinitely Many Solutions

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

Solving a System of Linear Equations with No Solution

Try it.

Solve the system by a method of your choice:

\[\{\begin{array}{lll}y & = & \frac{1}{2}x-3 \\ x-2y & = & 4\end{array}\]

Solution

Let us solve the system of linear equations by graphing.

\[\{\begin{array}{lll}y & = & \frac{1}{2}x-3 \\ x-2y & = & 4\end{array}\]

To graph the first equation, we will use its slope and $y$-intercept.

\[\begin{array}{lll}y & = & \frac{1}{2}x-3 \\ m & = & \frac{1}{2} \\ b & = & -3\end{array}\]

To graph the second equation, we will use the intercepts.

\[x-2y=4\]

$x$	$y$
0	−2
4	0

Graph the lines ().

Determine the points of intersection. The lines are parallel. Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system.

Solving a System of Linear Equations with Infinite Solutions

Try it.

Solve the system by a method of your choice:

$\{\begin{array}{lll}y & = & 2x-3 \\ -6x+3y & = & -9\end{array}$

Solution

Let us solve the system of linear equations by graphing.

$\{\begin{array}{lll}y & = & 2x-3 \\ -6x+3y & = & -9\end{array}$

Find the slope and $y$-intercept of the first equation.

$\begin{array}{lll}y & = & 2x-3 \\ m & = & 2 \\ b & = & -3\end{array}$

Find the intercepts of the second equation.

$-6x+3y=-9$

$x$	$y$
0	−3
$\frac{3}{2}$	0

Graph the lines ().

The lines are the same! 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 this system.

In the previous example, if you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same $y$-intercept. 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. We say the two lines are coincident. Coincident lines have the same slope and same $y$-intercept. A system of equations that has at least one solution is called a consistent system. A system with parallel lines has no solution. We call a system of equations like this an inconsistent system. It has no solution.

Lines	Intersecting	Parallel	Coincident
Number of Solutions	1 point	No solution	Infinitely many
Consistent/Inconsistent	Consistent	Inconsistent	Consistent
Dependent/Independent	Independent	Independent	Dependent

Condensed — the full section is in OpenStax Contemporary Mathematics.

Solving Applications of Systems of Linear Equations

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we will first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

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.

Applying System to a Real-World Application

Try it.

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 a year plus $15 for each training session. Option B would pay her $10,000 a year plus $40 for each training session. How many training sessions would make the salary options equal?

Solution

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for the number of training sessions that would make the pay equal.

Step 3: Name what we are looking for.

Let $s=\text{Heather’s salary}$, and $n=\text{the number of training sessions}$

Step 4: Translate into a system of equations.

Option A would pay her $25,000 plus $15 for each training session. \[s=25,000+15n\]

Option B would pay her $10,000 + $40 for each training session. \[s=10,000+40n\]

The system is shown. \[\{\begin{array}{l}s=25,000+15n \\ s=10,000+40n\end{array}\]

Step 5: Solve the system of equations.

We will use substitution.

Substitute $25,000+15n$ for $s$ in the second equation \[\begin{array}{l}s=25,000+15n \\ s=10,000+40n\end{array}\]

Solve for $n$. \[\begin{array}{lll}25,000+15n & = & 10,000+40n \\ 25,000 & = & 10,000+25n \\ 15,000 & = & 25n \\ 600 & = & n\end{array}\]

Step 6: Check the answer.

Are 600 training sessions a year reasonable?

Are the two options equal when $n=600$?

Substitute into each equation.

$\begin{array}{l}s=25,000+15(600)=34,000 \\ s=10,000+40(600)=34,000\end{array}$

Step 7: Answer the question.

The salary options would be equal for 600 training sessions.

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

To solve a system of linear equations means finding the point or points where the two linear equations intersect.
Two lines can intersect at one point, no points if they are parallel, or every point if they are the same equation.
Systems of linear equations can be solved by graphing, by using substitution, or by using the elimination method.

Practice (8)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

Determine whether the ordered pair is a solution to the system.
\[\{\begin{array}{lll}x-y & = & -1 \\ 2x-y & = & -5\end{array}\]
1. $(-2,-1)$
2. $(-4,-3)$
Odkrij odgovor
1. \[\{\begin{array}{lll}x-y & = & -1 \\ 2x-y & = & -5\end{array}\] We substitute $x=-2$ and $y=-1$ into both equations.
  \[\begin{array}{l}x-y=-1\ \\ -2-(-1)\overset{?}{=}-1\ \\ -1=-1✓\end{array}\ \begin{array}{l}2x-y=-5 \\ 2(-2)-(-1)\overset{?}{=}-5 \\ -3\ne -5\end{array}\]
  $(-2,-1)$ does not make both equations true. $(-2,-1)$ is not a solution.
2. We substitute $x=-4$ and $y=-3$ into both equations.
  \[\begin{array}{l}x-y=-1\ \\ -4-(-3)\overset{?}{=}-1\ \\ -1=-1✓\end{array}\ \begin{array}{l}2x-y=-5\ \\ 2•(-4)-(-3)\overset{?}{=}-5\ \\ -5=-5✓\end{array}\]
  $(-4,-3)$ makes both equations true. $(-4,-3)$ is a solution.
Determine whether the ordered pair is a solution to the system
\[\{\begin{array}{lll}y & = & \frac{3}{2}x+1 \\ 2x-3y & = & 7\end{array}\]
1. $(-4,-5)$
2. $(-4,5)$
Odkrij odgovor
1. \[\{\begin{array}{lll}y & = & \frac{3}{2}x+1 \\ 2x-3y & = & 7\end{array}\] Substitute $-4$ for $x$ and $-5$ for $y$ into both equations. \[\begin{array}{ll}\begin{array}{l}-5\overset{?}{=}\frac{3}{2}(-4)+1 \\ -5\overset{?}{=}3(-2)+1 \\ -5\overset{?}{=}-6+1 \\ -5=-5✓\end{array} & \begin{array}{l}2(-4)-3(-5)\overset{?}{=}7 \\ (-8)-(-15)\overset{?}{=}7 \\ -8+15\overset{?}{=}7 \\ 7=7✓\end{array}\end{array}\] $(-4,-5)$ is a solution.
2. \[\{\begin{array}{lll}y & = & \frac{3}{2}x+1 \\ 2x-3y & = & 7\end{array}\] Substitute $-4$ for $x$ and $5$ for $y$ into both equations. \[\begin{array}{l}5\overset{?}{=}\frac{3}{2}(-4)+1 \\ 5\overset{?}{=}3(-2)+1 \\ 5\overset{?}{=}-6+1 \\ 5\ne -5\end{array}\ \begin{array}{l}2(-4)-3(5)\overset{?}{=}7 \\ (-8)-(15)\overset{?}{=}7 \\ -8-15\overset{?}{=}7 \\ -23\ne 7\end{array}\] $(-4,5)$ is not a solution.

Solve this system of linear equations by graphing.

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

Odkrij odgovor

Step 1: Graph the first equation.	To graph the first line, write the equation in slope-intercept form. $\begin{array}{lllll}\ \begin{array}{lll}2x+y & = & 7 \\ y & = & -2x+7\end{array} \\ m=-2\ b=7\end{array}$	$\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}$
Step 2: Graph the second equation on the same rectangular coordinate system.	To graph the second line, use intercepts. $\begin{array}{l}x-2y=6 \\ (0,-3)\ (6,0)\end{array}$
Step 3: Determine whether the lines intersect, are parallel, or are the same line.	Look at the graph of the lines.	The lines intersect.
Step 4 and Step 5: Identify the solution to the system. 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.	Since the lines intersect, find the point of intersection. Check the point in both equations.	The lines intersect at $(4,-1)$. $\begin{array}{lll}2x+y & = & 7 \\ 2(4)+(-1) & \overset{?}{=} & 7 \\ 8-1 & \overset{?}{=} & 7 \\ 7 & = & 7✓ \\ & & \\ x-2y & = & 6 \\ 4-2(-1) & \overset{?}{=} & 6 \\ 6 & = & 6✓\end{array}$ The solution is $(4,-1)$.

Solve this system of linear equations by substitution:

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

Odkrij odgovor

Step 1: Solve one of the equations for either variable.	We’ll solve the first equation for $y$.	$\begin{array}{lllll}\begin{array}{l}\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\end{array} \\ \begin{array}{lll}2x+y & = & 7 \\ y & = & 7-2x\end{array}\end{array}$
Step 2: Substitute the expression from Step 1 into the other equation.	We replace $y$ in the second equation with the expression $7-2x$.	$\begin{array}{l}x-2y=6 \\ x-2(7-2x)=6\end{array}$
Step 3: Solve the resulting equation.	Now we have an equation with just 1 variable. We know how to solve this!	$\begin{array}{lll}x-2(7-2x) & = & 6 \\ x-14+4x & = & 6 \\ 5x & = & 20 \\ x & = & 4\end{array}$
Step 4: Substitute the solution from Step 3 into one of the original equations to find the other variable.	We’ll use the first equation and replace $x$ with 4.	$\begin{array}{lll}2x+y & = & 7 \\ 2(4)+y & = & 7 \\ 8+y & = & 7 \\ y & = & -1\end{array}$
Step 5: Write the solution as an ordered pair.	The ordered pair is ($x$, $y$).	$(4,-1)$
Step 6: Check that the ordered pair is a solution to both original equations.	Substitute $x=4$, $y=-1$ into both equations and make sure they are both true.	$\begin{array}{llllllllllllll}\begin{array}{lll}2x+y & = & 7 \\ 2(4)+(-1) & \overset{?}{=} & 7 \\ 7 & = & 7✓\end{array} & \begin{array}{lll}x-2y & = & 6 \\ 4-2(-1) & \overset{?}{=} & 6 \\ 6 & = & 6✓\end{array}\end{array}$ Both equations are true. $(4,-1)$ is the solution to the system.

Solve this system of linear equations by elimination:

\[\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\]

Odkrij odgovor

Step 1: Write both equations in standard form. If any coefficients are fractions, clear them.	Both equations are in standard form, $Ax+By=C$. There are no fractions.	$\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}$
Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.	We can eliminate the $y$’s by multiplying the first equation by 2. Multiply both sides of $2x+y=7$ by 2.	$\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}$ $\{\begin{array}{lll}2(2x+y) & = & 2(7) \\ x-2y & = & 6\end{array}$
Step 3: Add the equations resulting from Step 2 to eliminate one variable.	We add the $x$’s, $y$’s, and constants.	$\begin{array}{l}\{\begin{array}{l}4x+2y=14 \\ \underset{_}{x-2y=6\ }\end{array} \\ 5x\ =20\ \end{array}$
Step 4: Solve for the remaining variable.	Solve for $x$.	$x=4$
Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.	Substitute $x=4$ into the second equation, $x-2y=6$. Then solve for $y$.	$\begin{array}{lll}x-2y & = & 6 \\ 4-2y & = & 6 \\ -2y & = & 2 \\ y & = & -1\end{array}$
Step 6: Write the solution as an ordered pair.	Write it as $(x,y)$.	$(4,-1)$
Step 7: Check that the ordered pair is a solution to both original equations.	Substitute $x=4$, $y=-1$ into $2x+y=7$ and $x-2y=6$. Do they make both equations true? Yes!	$\begin{array}{llllllllllllll}\begin{array}{lll}2x+y & = & 7 \\ 2(4)+(-1) & \overset{?}{=} & 7 \\ 7 & = & 7✓\end{array} & \begin{array}{lll}x-2y & = & 6 \\ 4-2(-1) & \overset{?}{=} & 6 \\ 6 & = & 6✓\end{array}\end{array}$ The solution is $(4,-1)$ .

Solve the system by a method of your choice:
\[\{\begin{array}{lll}y & = & \frac{1}{2}x-3 \\ x-2y & = & 4\end{array}\]

Odkrij odgovor

Let us solve the system of linear equations by graphing.
\[\{\begin{array}{lll}y & = & \frac{1}{2}x-3 \\ x-2y & = & 4\end{array}\]
To graph the first equation, we will use its slope and $y$-intercept.
\[\begin{array}{lll}y & = & \frac{1}{2}x-3 \\ m & = & \frac{1}{2} \\ b & = & -3\end{array}\]
To graph the second equation, we will use the intercepts.
\[x-2y=4\]
$x$ $y$
0 −2
4 0
Graph the lines ().
Determine the points of intersection. The lines are parallel. Since no point is on both lines, there is no ordered pair that makes both equations true. There is no solution to this system.
Solve the system by a method of your choice:
$\{\begin{array}{lll}y & = & 2x-3 \\ -6x+3y & = & -9\end{array}$

Odkrij odgovor

Let us solve the system of linear equations by graphing.
$\{\begin{array}{lll}y & = & 2x-3 \\ -6x+3y & = & -9\end{array}$
Find the slope and $y$-intercept of the first equation.
$\begin{array}{lll}y & = & 2x-3 \\ m & = & 2 \\ b & = & -3\end{array}$
Find the intercepts of the second equation.
$-6x+3y=-9$
$x$ $y$
0 −3
$\frac{3}{2}$ 0
Graph the lines ().
The lines are the same! 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 this system.
Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $25,000 a year plus $15 for each training session. Option B would pay her $10,000 a year plus $40 for each training session. How many training sessions would make the salary options equal?

Odkrij odgovor

Step 1: Read the problem.
Step 2: Identify what we are looking for.
We are looking for the number of training sessions that would make the pay equal.
Step 3: Name what we are looking for.
Let $s=\text{Heather’s salary}$, and $n=\text{the number of training sessions}$
Step 4: Translate into a system of equations.
Option A would pay her $25,000 plus $15 for each training session. \[s=25,000+15n\]
Option B would pay her $10,000 + $40 for each training session. \[s=10,000+40n\]
The system is shown. \[\{\begin{array}{l}s=25,000+15n \\ s=10,000+40n\end{array}\]
Step 5: Solve the system of equations.
We will use substitution.
Substitute $25,000+15n$ for $s$ in the second equation \[\begin{array}{l}s=25,000+15n \\ s=10,000+40n\end{array}\]
Solve for $n$. \[\begin{array}{lll}25,000+15n & = & 10,000+40n \\ 25,000 & = & 10,000+25n \\ 15,000 & = & 25n \\ 600 & = & n\end{array}\]
Step 6: Check the answer.
Are 600 training sessions a year reasonable?
Are the two options equal when $n=600$?
Substitute into each equation.
$\begin{array}{l}s=25,000+15(600)=34,000 \\ s=10,000+40(600)=34,000\end{array}$
Step 7: Answer the question.
The salary options would be equal for 600 training sessions.

Symbols used here

\neq

not equal
The two sides are different.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

|x|

absolute value / modulus
Distance from zero: |−3| = 3. For a complex number, distance from the origin.

imaginary unit
i² = −1.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\mathbb{N},\ \mathbb{Z},\ \mathbb{Q},\ \mathbb{R},\ \mathbb{C}

number sets
Naturals, integers, rationals, reals, complex numbers.

How to: Systems of Linear Equations in Two Variables

Determine and show whether an ordered pair is a solution to a system of equations.
Solve systems of linear equations using graphical methods.
Solve systems of linear equations using substitution.
Solve systems of linear equations using elimination.
Identify systems with no solution or infinitely many solutions.
Solve applications of systems of linear equations.
We substitute

Questions people ask

What does it mean to solve an equation?

To find every value of the unknown that makes both sides equal. Each step is an operation applied to both sides that keeps the solution set the same, until the unknown stands alone.

Why do I sometimes get two answers?

A quadratic can cross the axis twice, so it can have two solutions. A degree-n polynomial has up to n. The graph shows where each one comes from.

How do I know whether to factor or use the quadratic formula?

Try factoring for a few seconds: look for two numbers that multiply to a·c and add to b. If nothing obvious appears, the discriminant b² − 4ac tells you how many real roots there are, and the formula finds them without guessing.

Poskusi sam.

Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

Več v Algebra

Linear equations Quadratic equations Systems of equations Inequalities Factoring Expanding Simplifying expressions Functions and graphs Exponential and logarithmic equations Polynomial equations Absolute value

Step 1: Solve one of the equations for either variable.	We’ll solve the first equation for \(y\).	\(\begin{array}{lllll}\begin{array}{l}\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\end{array} \\ \begin{array}{lll}2x+y & = & 7 \\ y & = & 7-2x\end{array}\end{array}\)
Step 2: Substitute the expression from Step 1 into the other equation.	We replace \(y\) in the second equation with the expression \(7-2x\).	\(\begin{array}{l}x-2y=6 \\ x-2(7-2x)=6\end{array}\)
Step 3: Solve the resulting equation.	Now we have an equation with just 1 variable. We know how to solve this!	\(\begin{array}{lll}x-2(7-2x) & = & 6 \\ x-14+4x & = & 6 \\ 5x & = & 20 \\ x & = & 4\end{array}\)
Step 4: Substitute the solution from Step 3 into one of the original equations to find the other variable.	We’ll use the first equation and replace \(x\) with 4.	\(\begin{array}{lll}2x+y & = & 7 \\ 2(4)+y & = & 7 \\ 8+y & = & 7 \\ y & = & -1\end{array}\)
Step 5: Write the solution as an ordered pair.	The ordered pair is (\(x\), \(y\)).	\((4,-1)\)
Step 6: Check that the ordered pair is a solution to both original equations.	Substitute \(x=4\), \(y=-1\) into both equations and make sure they are both true.	\(\begin{array}{llllllllllllll}\begin{array}{lll}2x+y & = & 7 \\ 2(4)+(-1) & \overset{?}{=} & 7 \\ 7 & = & 7✓\end{array} & \begin{array}{lll}x-2y & = & 6 \\ 4-2(-1) & \overset{?}{=} & 6 \\ 6 & = & 6✓\end{array}\end{array}\) Both equations are true. \((4,-1)\) is the solution to the system.

\(x\)	\(y\)
0	−3
\(\frac{3}{2}\)	0

Step 1: Solve one of the equations for either variable.	We’ll solve the first equation for \(y\).	\(\begin{array}{lllll}\begin{array}{l}\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\end{array} \\ \begin{array}{lll}2x+y & = & 7 \\ y & = & 7-2x\end{array}\end{array}\)
Step 2: Substitute the expression from Step 1 into the other equation.	We replace \(y\) in the second equation with the expression \(7-2x\).	\(\begin{array}{l}x-2y=6 \\ x-2(7-2x)=6\end{array}\)
Step 3: Solve the resulting equation.	Now we have an equation with just 1 variable. We know how to solve this!	\(\begin{array}{lll}x-2(7-2x) & = & 6 \\ x-14+4x & = & 6 \\ 5x & = & 20 \\ x & = & 4\end{array}\)
Step 4: Substitute the solution from Step 3 into one of the original equations to find the other variable.	We’ll use the first equation and replace \(x\) with 4.	\(\begin{array}{lll}2x+y & = & 7 \\ 2(4)+y & = & 7 \\ 8+y & = & 7 \\ y & = & -1\end{array}\)
Step 5: Write the solution as an ordered pair.	The ordered pair is (\(x\), \(y\)).	\((4,-1)\)
Step 6: Check that the ordered pair is a solution to both original equations.	Substitute \(x=4\), \(y=-1\) into both equations and make sure they are both true.	\(\begin{array}{llllllllllllll}\begin{array}{lll}2x+y & = & 7 \\ 2(4)+(-1) & \overset{?}{=} & 7 \\ 7 & = & 7✓\end{array} & \begin{array}{lll}x-2y & = & 6 \\ 4-2(-1) & \overset{?}{=} & 6 \\ 6 & = & 6✓\end{array}\end{array}\) Both equations are true. \((4,-1)\) is the solution to the system.

Step 1: Write both equations in standard form. If any coefficients are fractions, clear them.	Both equations are in standard form, \(Ax+By=C\). There are no fractions.	\(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\)
Step 2: Make the coefficients of one variable opposites. Decide which variable you will eliminate. Multiply one or both equations so that the coefficients of that variable are opposites.	We can eliminate the \(y\)’s by multiplying the first equation by 2. Multiply both sides of \(2x+y=7\) by 2.	\(\{\begin{array}{l}2x+y=7 \\ x-2y=6\end{array}\) \(\{\begin{array}{lll}2(2x+y) & = & 2(7) \\ x-2y & = & 6\end{array}\)
Step 3: Add the equations resulting from Step 2 to eliminate one variable.	We add the \(x\)’s, \(y\)’s, and constants.	\(\begin{array}{l}\{\begin{array}{l}4x+2y=14 \\ \underset{_}{x-2y=6\ }\end{array} \\ 5x\ =20\ \end{array}\)
Step 4: Solve for the remaining variable.	Solve for \(x\).	\(x=4\)
Step 5: Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.	Substitute \(x=4\) into the second equation, \(x-2y=6\). Then solve for \(y\).	\(\begin{array}{lll}x-2y & = & 6 \\ 4-2y & = & 6 \\ -2y & = & 2 \\ y & = & -1\end{array}\)
Step 6: Write the solution as an ordered pair.	Write it as \((x,y)\).	\((4,-1)\)
Step 7: Check that the ordered pair is a solution to both original equations.	Substitute \(x=4\), \(y=-1\) into \(2x+y=7\) and \(x-2y=6\). Do they make both equations true? Yes!	\(\begin{array}{llllllllllllll}\begin{array}{lll}2x+y & = & 7 \\ 2(4)+(-1) & \overset{?}{=} & 7 \\ 7 & = & 7✓\end{array} & \begin{array}{lll}x-2y & = & 6 \\ 4-2(-1) & \overset{?}{=} & 6 \\ 6 & = & 6✓\end{array}\end{array}\) The solution is \((4,-1)\) .

\(x\)	\(y\)
0	−3
\(\frac{3}{2}\)	0