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 3.2.
Find the slope and y-intercept of the line 3 x − y = 12 . 3 x − y = 12 .
If you missed this problem, review Example 3.16.
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 3.8.
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.
< 2 x + y = 7 x − 2 y = 6 < 2 x + y = 7 x − 2 y = 6A linear equation in two variables, such as 2 x + y = 7 , 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 ) ( x , y ) that make both equations true. These are called the solutions of a system of equations .
The 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 ) . ( 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.
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 )
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.
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.
We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y-intercepts.
| Solve the first equation for y. | |
| Find the slope and y-intercept. | |
| Solve the second equation for y. | |
| Find the slope and y-intercept. | |
| Graph the lines. | |
| Determine the point of intersection. | The lines intersect at ( −1 , 2 ) . ( −1 , 2 ) . |
| Check the solution in both equations. | |
| The solution is ( −1 , 2 ) . ( −1 , 2 ) . |
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.
| To graph the first equation, we will use its slope and y-intercept. | |
| To graph the second equation, we will use the intercepts. | |
| 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. |
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.
| Find the slope and y-intercept of the first equation. |
| Find the intercepts of the second equation. |
| 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. |
If you write the second equation 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 have the same slope and same y-intercept.
The systems of equations in Example 4.2 and Example 4.3 each had two intersecting lines. Each system had one solution.
In Example 4.5, 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 Example 4.4, 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 Table 4.1.
| Lines | Intersecting | Parallel | Coincident |
|---|---|---|---|
| Number of solutions | 1 point | No solution | Infinitely many |
| Consistent/inconsistent | Consistent | Inconsistent | Consistent |
| Dependent/ independent | Independent | Independent | Dependent |
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.
| The first equation is already in slope-intercept form. | < y = 3 x − 1 6 x − 2 y = 12 y = 3 x − 1 < y = 3 x − 1 6 x − 2 y = 12 y = 3 x − 1 |
| Write the second equation in slope-intercept form. | 6 x − 2 y = 12 − 2 y = −6 x + 12 −2 y −2 = −6 x + 12 −2 y = 3 x − 6 6 x − 2 y = 12 − 2 y = −6 x + 12 −2 y −2 = −6 x + 12 −2 y = 3 x − 6 |
| Find the slope and intercept of each line. | y = 3 x − 1 y = 3 x − 6 m = 3 m = 3 b = −1 b = −6 y = 3 x − 1 y = 3 x − 6 m = 3 m = 3 b = −1 b = −6 |
| Since the slopes are the same and y-intercepts are different, the lines are parallel. |
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.
| < 2 x + y = −3 x − 5 y = 5 < 2 x + y = −3 x − 5 y = 5 | |
| Write both equations in slope–intercept form. | 2 x + y = − 3 x − 5 y = 5 y = −2 x − 3 −5 y = − x + 5 − 5 y − 5 = − x + 5 − 5 y = 1 5 x − 1 2 x + y = − 3 x − 5 y = 5 y = −2 x − 3 −5 y = − x + 5 − 5 y − 5 = − x + 5 − 5 y = 1 5 x − 1 |
| Find the slope and intercept of each line. | y = − 2 x − 3 y = 1 5 x − 1 m = − 2 m = 1 5 b = −3 b = −1 y = − 2 x − 3 y = 1 5 x − 1 m = − 2 m = 1 5 b = −3 b = −1 |
| Since the slopes are different, the lines intersect. |
A system of equations whose graphs are intersect has 1 solution and is consistent and independent.
Without graphing, determine the number of solutions and then classify the system of equations.
Without graphing, determine the number of solutions and then classify the system of equations.
Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.
We will now solve systems of linear equations by the substitution method.
We will use the same system we used first for graphing.
< 2 x + y = 7 x − 2 y = 6 < 2 x + y = 7 x − 2 y = 6We 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 the system by substitution: < 2 x + y = 7 x − 2 y = 6 . < 2 x + y = 7 x − 2 y = 6 .
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 the first equation for y. Substitute − 2 x + 2 − 2 x + 2 for y in the second equation. |
| Replace the y with − 2 x + 2 . − 2 x + 2 . |
| Solve the equation for x. |
| Substitute x = 5 4 x = 5 4 into 4 x + 2 y = 4 4 x + 2 y = 4 to find y. |
| The ordered pair is ( 5 4 , − 1 2 ) . ( 5 4 , − 1 2 ) . |
| Check the ordered pair in both equations. |
| The solution is ( 5 4 , − 1 2 ) . ( 5 4 , − 1 2 ) . |
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 . 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:
< 3 x + y = 5 2 x − y = 0 ————— 5 x = 5 < 3 x + y = 5 2 x − y = 0 ————— 5 x = 5The y’s add to zero and we have one equation with one variable.
Let’s try another one:
< x + 4 y = 2 2 x + 5 y = −2 < x + 4 y = 2 2 x + 5 y = −2This time we don’t see a variable that can be immediately eliminated if we add the equations.
But if we multiply the first equation by −2 , −2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2 . −2 .
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.
