Using Graphs and Tables to Solve Linear Systems 3-1 Warm Up Lesson Presentation Lesson Quiz Holt Algebra 2

Warm Up Use substitution to determine if (1, –2) is an element of the solution set of the linear equation. no yes 1. y = 2x + 1 2. y = 3x – 5 Write each equation in slope-intercept form. 4. 4y – 3x = 8 3. 2y + 8x = 6 y = –4x + 3

Objectives Solve systems of equations by using graphs and tables. Classify systems of equations, and determine the number of solutions.

Vocabulary system of equations linear system consistent system inconsistent system independent system dependent system

A system of equations is a set of two or more equations containing two or more variables. A linear system is a system of equations containing only linear equations.

On the graph of the system of two equations, the solution is the set of points where the lines intersect. A point is a solution to a system of equation if the x- and y-values of the point satisfy both equations.

Example 1A: Verifying Solutions of Linear Systems Use substitution to determine if the given ordered pair is an element of the solution set for the system of equations. (1, 3); x – 3y = –8 3x + 2y = 9 x – 3y = –8 (1) –3(3) –8  3x + 2y = 9 3(1) +2(3) 9  Substitute 1 for x and 3 for y in each equation. Because the point is a solution for both equations, it is a solution of the system.

Example 1B: Verifying Solutions of Linear Systems Use substitution to determine if the given ordered pair is an element of the solution set for the system of equations. x + 6 = 4y (–4, ); 2x + 8y = 1 x + 6 = 4y (–4) + 6 2 2x + 8y = 1 2(–4) + 1 –4 Substitute –4 for x and for y in each equation.  x Because the point is not a solution for both equations, it is not a solution of the system.

Example 2A: Solving Linear Systems by Using Graphs and Tables Use a graph and a table to solve the system. Check your answer. 2x – 3y = 3 y + 2 = x y= x – 2 y= x – 1 Solve each equation for y.

Example 2A Continued On the graph, the lines appear to intersect at the ordered pair (3, 1)

1 3 2 –1 y x x y –2 1 – 1 2 3 Example 2A Continued y= x – 1 y= x – 2 y x x y –2 1 – 1 2 3 Make a table of values for each equation. Notice that when x = 3, the y-value for both equations is 1. The solution to the system is (3, 1).

The systems of equations in Example 2 have exactly one solution The systems of equations in Example 2 have exactly one solution. However, linear systems may also have infinitely many or no solutions. A consistent system is a set of equations or inequalities that has at least one solution, and an inconsistent system will have no solutions.

You can classify linear systems by comparing the slopes and y-intercepts of the equations. An independent system has equations with different slopes. A dependent system has equations with equal slopes and equal y-intercepts.

Example 3A: Classifying Linear System Classify the system and determine the number of solutions. x = 2y + 6 3x – 6y = 18 y = x – 3 The equations have the same slope and y-intercept and are graphed as the same line. Solve each equation for y. The system is consistent and dependent with infinitely many solutions.

Example 3B: Classifying Linear System Classify the system and determine the number of solutions. 4x + y = 1 y + 1 = –4x The equations have the same slope but different y-intercepts and are graphed as parallel lines. y = –4x + 1 y = –4x – 1 Solve each equation for y. The system is inconsistent and has no solution.

yes no 3. Lesson Quiz: Part I Use substitution to determine if the given ordered pair is an element of the solution set of the system of equations. x + 3y = –9 x + y = 2 2. (–3, –2) 1. (4, –2) y – 2x = 4 y + 2x = 5 yes no Solve the system using a table and graph. Check your answer. x + y = 1 (2, –1) 3. 3x –2y = 8

Lesson Quiz: Part II Classify each system and determine the number of solutions. –4x = 2y – 10 y + 2x = –10 4. 5. y + 2x = –10 y + 2x = –10 consistent, dependent; infinitely many inconsistent; none 6. Kayak Kottage charges $26 to rent a kayak plus $24 per hour for lessons. Power Paddles charges $12 for rental plus $32 per hour for lessons. For what number of hours is the cost of equipment and lessons the same for each company?