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5.1 Solve systems of equations by graphing  (Page 15/14)

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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

Before you get started, take this readiness quiz.

  1. For the equation y = 2 3 x − 4
    ⓐ is ( 6 , 0 ) a solution? ⓑ is ( −3 , −2 ) a solution?
    If you missed this problem, review [link] .
  2. Find the slope and y -intercept of the line 3 x − y = 12 .
    If you missed this problem, review [link] .
  3. Find the x - and y -intercepts of the line 2 x − 3 y = 12 .
    If you missed this problem, review [link] .

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.

{ 2 x + y = 7 x − 2 y = 6

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:

{ 3 x − y = 7 x − 2 y = 4

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

This figure begins with a sentence, “We substitute x =2 and y = -1 into both equations.” The first equation shows that 3x minus y equals 7. Then 3 times 2 minus negative, in parentheses, equals 7. Then 7 equals 7 is true. The second equation reads x minus 2y equals 4. Then 2 minus 2 times negative one in parentheses equals 4. Then 4 = 4 is true.

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?

This figure begins with the sentence, “We substitute x equals 3 and y equals 2 into both equations.” The first equation reads 3 times x minus 7equals 7. Then, 3 times 3 minus 2 equals 7. Then 7 = 7 is true. The second equation reads x minus 2y equals 4. The n times 2 minus 2 times 2 = 4. Then negative 2 = 4 is false.

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.

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

ⓐ ( −2 , −1 ) ⓑ ( −4 , −3 )

Solution

  1. ⓐ
    This figure shows two bracketed equations. The first is x minus y = negative 1. The second is 2 times x minus y equals negative 5. The sentence, “We substitute x = negative 2 and y = 1 into both equations,” follows. The first equation shows the substitution and reveals that negative 1 = negative 1. The second equation shows the substitution and reveals that 5 do not equal -5. Under the first equation is the sentence, “(negative 2, negative 1) does not make both equations true.” Under the second equation is the sentence, “(negative 2, negative 1) is not a solution.”
    (−2, −1) does not make both equations true. (−2, −1) is not a solution.

    ⓑ
    This figure begins with the sentence, “We substitute x = -4 and y = -3 into both equations.” The first equation listed shows x – y = -1. Then -4 - (-3) = -1. Then -1 = -1. The second equation listed shows 2x – y = -5. Then 2 times (-4) – (-3) = -5. Then -5 = -5. Under the first equation is the sentence, “(-4, -3) does make both equations true.” Under the second equation is the sentence, “(-4, -3) is a solution.”
    (−4, −3) does not make both equations true. (−4, −3) is a solution.
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Source:  OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2
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