1. Solving Systems of Equations in Two Variables; Applications
Section 12.1
Solving Systems of Equations in Two Variables; Applications

2. Objectives Determine whether an ordered pair is a solution of a system
Solve systems of linear equations by graphing
Use graphing to identify inconsistent systems and dependent equations
Solve systems of linear equations by substitution
Solve systems of linear equations by the elimination (addition) method
Use substitution and elimination (addition) to identify inconsistent systems and dependent equations
Solve application problems using systems of equations

3. Objective 1: Determine Whether an Ordered Pair Is a Solution of a System
When two equations with the same variables are considered simultaneously (at the same time), we say that they form a system of equations. We will use a left brace { when writing a system of equations. An example is
A solution of a system of equations in two variables is an ordered pair that satisfies both equations of the system.

4. EXAMPLE 1
Determine whether (−3, 1) is a solution of each system of equations.
Strategy We will substitute the x- and y-coordinates of (−3, 1) for the corresponding variables in both equations of the system.
Why If both equations are satisfied (made true) by the x- and y-coordinates, the ordered pair is a solution of the system.

5. EXAMPLE 1
Determine whether (−3, 1) is a solution of each system of equations.
Solution
a. To determine whether (−3, 1) is a solution, we substitute −3 for x and 1 for y in each equation.
Since (−3, 1) satisfies both equations, it is a solution of the system.

6. EXAMPLE 1
Determine whether (−3, 1) is a solution of each system of equations.
Solution
b. Again, we substitute −3 for x and 1 for y in each equation in the second system .
Although (−3, 1) satisfies the first equation, it does not satisfy the second. Because it does not satisfy both equations, (−3, 1) is not a solution of the system.

7. Objective 2: Solve Systems of Linear Equations by Graphing
To solve a system of linear equations in two variables by graphing, we use the following steps.
Carefully graph each equation on the same rectangular coordinate system.
If the lines intersect, determine the coordinates of the point of intersection of the graphs. That ordered pair is the solution of the system.
If the graphs have no point in common, the system has no solution.
Check the proposed solution in each equation of the original system.
When a system of equations (as in Example 2) has at least one solution, the system is called a consistent system.

8. EXAMPLE 2
Solve the system by graphing:
Strategy We will graph both equations on the same coordinate system.
Why The graph of a linear equation is a picture of its solutions. If both equations are graphed on the same coordinate system, we can see whether they have any common solutions.

9. EXAMPLE 2 Solution Solve the system by graphing:
The intercept method is a convenient way to graph equations such as x + 2y = 4 and 2x – y = 3, because they are in standard Ax + By = C form.

10. EXAMPLE 2 Solution Solve the system by graphing:
Although infinitely many ordered pairs (x, y) satisfy x + 2y = 4 and infinitely many ordered pairs (x, y) satisfy 2x – y = 3, only the coordinates of the point where the graphs intersect satisfy both equations. From the graph, it appears that the intersection point has coordinates (2, 1). To verify that it is the solution, we substitute 2 for x and 1 for y in both equations and show that (2, 1) satisfies each one.
Since (2, 1) makes both equations true, it is the solution of the system. The solution set is {(2, 1)}.

11. Objective 3: Use Graphing to Identify Inconsistent Systems and Dependent Equations
When a system has no solution (as in Example 3), it is called an inconsistent system.
When the equations of a system have different graphs (as in Examples 2 and 3), the equations are called independent equations. Two equations with the same graph are called dependent equations.
If each equation in one system is equivalent to a corresponding equation in another system, the systems are called equivalent systems.

12. Objective 3: Use Graphing to Identify Inconsistent Systems and Dependent Equations
If the lines are different and intersect, the equations are independent, and the system is consistent. One solution exists. It is the point of intersection.
If the lines are different and parallel, the equations are independent, and the system is inconsistent. No solution exists.
If the lines coincide, the equations are dependent, and the system is consistent. Infinitely many solutions exist. Any point on the line is a solution.

13. EXAMPLE 3
Solve the system by graphing, if possible:
Strategy We will graph both equations on the same coordinate system.
Why If both equations are graphed on the same coordinate system, we can see whether they have any common solutions.

14. EXAMPLE 3
Solve the system by graphing, if possible:
Solution
Using the intercept method, we graph both equations on one set of coordinate axes, as shown below.

15. EXAMPLE 3
Solve the system by graphing, if possible:
Solution
In this example, the graphs are parallel because the slopes of the lines are equal and they have different y-intercepts. We can see that the slope of each line is by writing each equation in slope-intercept form. To do that, we solve each for y.
Because the lines are parallel, there is no point of intersection. Such a system has no solution and it is called an inconsistent system. The solution set is the empty set, which is written Ø.

16. Objective 4: Solve Systems of Linear Equations by Substitution
The substitution method works well for solving systems where one equation is solved, or can be easily solved, for one of the variables. To solve a system of two linear equations in x and y by the substitution method, we can follow these steps.
Solve one of the equations for either x or y—preferably a variable with a coefficient of 1 or −1. If this is already done, go to step 2. (We call this equation the substitution equation.)
Substitute the expression for x or for y obtained in step 1 into the other equation and solve that equation.
Substitute the value of the variable found in step 2 into the substitution equation to find the value of the remaining variable.
Check the proposed solution in each equation of the original system. Write the solution as an ordered pair.

17. EXAMPLE 6
Solve the system by substitution:
Strategy We will use the substitution method. Since the system does not contain an equation solved for x or y, we must choose an equation and solve it for x or y. It is easiest to solve for y in the first equation, because y has a coefficient of 1.
Why Solving 4x + y = 13 for x or solving
–2x + 3y = –17 for x or y would involve working with cumbersome fractions.

18. EXAMPLE 6
Solve the system by substitution:
Solution
Step 1: We solve the first equation for y, because y has a coefficient of 1.
Because y and –4x + 13 are equal, we can substitute –4x + 13 for y in the second equation of the system.

19. EXAMPLE 6
Solve the system by substitution:
Solution
Step 2: We then substitute –4x + 13 for y in the second equation to eliminate the variable y from that equation. The result will be an equation containing only one variable, x.

20. EXAMPLE 6
Solve the system by substitution:
Solution
Step 3: To find y, we substitute 4 for x in the substitution equation and evaluate the right side.
Multiply: –4(4) = –16.

21. EXAMPLE 6
Solve the system by substitution:
Solution
Step 4: To verify that (4, −3) satisfies both equations, we substitute 4 for x and −3 for y into each equation of the original system and simplify.
Since (4, −3) satisfies both equations of the system, it is the solution of the system. The solution set is {(4, −3)}. The graphs of the equations of the system help to verify this—they appear to intersect at (4, −3), as shown on the left.

22. Objective 5: Solve Systems of Linear Equations by the Elimination (Addition) Method
With the elimination (addition) method, we combine the equations of the system in a way that will eliminate terms involving one of the variables.
Write both equations of the system in standard (general) form: Ax + By = C.
If necessary, multiply one or both of the equations by a nonzero number chosen to make the coefficients of x (or the coefficients of y) opposites.
Add the equations to eliminate the terms involving x (or y)
Solve the equation resulting from step 3.
Find the value of the remaining variable by substituting the solution found in step 4 into any equation containing both variables. Or, repeat steps 2–4 to eliminate the other variable.
Check the proposed solution in each equation of the original system. Write the solution as an ordered pair.

23. EXAMPLE 7
Solve:
Strategy We will find an equivalent system without fractions or decimals and use the elimination method to solve it.
Why It’s usually easier to solve a system of equations that involves only integers.

24. EXAMPLE 7
Solve:
Solution
Step 1: To clear the first equation of the fractions, we multiply both sides by 6. To clear the second equation of decimals, we multiply both sides by 10.

25. EXAMPLE 7
Solve:
Solution
Step 2: To make the y-terms drop out when we add the equations, we multiply both sides of 8x + 3y = –4 by 4 and both sides of 3x + 4y = 10 by –3 to get

26. EXAMPLE 7
Solve:
Solution
Step 3: When these equations are added, the y-terms drop out.
Step 4: We solve the resulting equation to find x.

27. EXAMPLE 7
Solve:
Solution
Step 5: To find y, we can substitute –2 for x in either of the equations of the original system or either of the equations of the equivalent system. It appears the computations will be the simplest if we use 3x + 4y = 10.
Step 6: The solution is (–2, 4) and the solution set is {(–2, 4)}. Verify that the solution checks using the original equations.

28. Objective 6: Use Substitution and Elimination (Addition) to Identify Inconsistent Systems and Dependent Equations
We have solved inconsistent systems and systems of dependent equations by graphing. We also can solve these systems using the substitution and elimination methods.
If the variables drop out and a true statement (identity) is obtained, the system has an infinite number of solutions. The equations are dependent and the system is consistent.
If the variables drop out and a false statement (contradiction) is obtained, the system has no solution and is inconsistent.

29. EXAMPLE 8
Solve:
Strategy We will use the substitution method to solve this system.
Why The substitution method works well when one of the equations of the system (in this case, y = 2x + 4) is solved for a variable.

30. EXAMPLE 8
Solve:
Solution
Since the first equation is solved for y we will use the substitution method.
We can now try to solve this equation for x:

31. EXAMPLE 8 Solution Solve:
Here, the terms involving x drop out, and we get –16 = 7. This false statement indicates that the system has no solution and is, therefore, inconsistent. The solution set is Ø. The graphs of the equations of the system help to verify this—they appear to be parallel lines, as shown below.

32. Objective 7: Solve Application Problems Using Systems of Equations
We have solved applied problems involving two unknown quantities by modeling the situation with an equation in one variable.
It’s often easier to solve such problems using a two-variable
approach.
We write two equations in two variables to model the situation, and then we use the methods of this section to solve the system formed by the pair of equations.

33. EXAMPLE 10 Analyze the Problem Assign
Wedding Pictures. A professional photographer offers two different packages
for wedding pictures. Use the information in the illustration given to determine the cost of one 8  10-inch photograph and the cost of one 5  7-inch photograph.
Analyze the Problem
■ Eight 8  10 and twelve 5  7 pictures cost $133.
■ Six 8  10 and twenty-two 5  7 pictures cost $168.
■ Find the cost of one 8  10 photograph and the cost of one 5  7 photograph.
Assign
Let x = the cost of one 8  10 photograph (in dollars), and let y = the cost of one 5  7 photograph (in dollars).

34. EXAMPLE 10 Form Two Equations
Wedding Pictures. A professional photographer offers two different packages
for wedding pictures. Use the information in the illustration given to determine the cost of one 8  10-inch photograph and the cost of one 5  7-inch photograph.
Form Two Equations
We can use the fact that Number  value = total value to construct tables that model the cost of each package.

35. EXAMPLE 10 Form Two Equations
Wedding Pictures. A professional photographer offers two different packages
for wedding pictures. Use the information in the illustration given to determine the cost of one 8  10-inch photograph and the cost of one 5  7-inch photograph.
Form Two Equations

36. EXAMPLE 10 Solve the System
Wedding Pictures. A professional photographer offers two different packages
for wedding pictures. Use the information in the illustration given to determine the cost of one 8  10-inch photograph and the cost of one 5  7-inch photograph.
Solve the System
We will use elimination to solve this system. To make the x-terms drop out, we multiply both sides of equation 1 by 3. Then we multiply both sides of equation 2 by –4, add the resulting equations, and solve for y:

37. EXAMPLE 10 Solve the System
Wedding Pictures. A professional photographer offers two different packages
for wedding pictures. Use the information in the illustration given to determine the cost of one 8  10-inch photograph and the cost of one 5  7-inch photograph.
Solve the System
To find x we substitute 5.25 for y in equation 1 and solve for x:

38. EXAMPLE 10 State the Conclusion Check the Result
Wedding Pictures. A professional photographer offers two different packages
for wedding pictures. Use the information in the illustration given to determine the cost of one 8  10-inch photograph and the cost of one 5  7-inch photograph.
State the Conclusion
The cost of one 8  10 photo is $8.75, and the cost of one photo 5  7 is $5.25.
Check the Result
If the first package contains eight 8  10 and twelve 5  7 photographs, the value of the package is 8($8.75) + 12($5.25) = $70 + $63 = $133.
If the second package contains six 8  10 and twenty-two 5  7 photographs, the value of the package is
6($8.75) + 22($5.25) = $ $ = $168. The results check.