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Bittinger Pre-Algebra, 6th edition Section 6.4 – Graphing Linear Equations Tom Atwater Draft 1 Example from Text: Exercises from Text: 7, 17, 37, 51, 61, 65 Student: No Video Audio [PP 1] Section 6.4 [PP 2] Graphing Linear Equations FULL SCREEN PRESENTER [PP 3] INTRODUCTION: Hi, my name is Tom. Today we are going to learn how to graph linear equations. READ Objectives. Objectives Find solutions of equations in two variables. Graph linear equations in two variables. Graph equations for horizontal or vertical lines. FULL SCREEN PRESENTER [PP 4] SAY: To solve an equation with two variables, we will first replace one variable with some number choice and then solve the resulting equation. Let’s work two examples. READ Exercise 7 For the equation, use the indicated value to find an ordered pair that is a solution. x + 3y = 1; let x = 10 MATH STEPS: x + 3y = 1; let x = 10 Substitute 10 for x and solve for y 10 + 3y = 1 3y = –9 y=–3 The ordered pair (10, –3) is a solution of 1 [PP 5] Bittinger Pre-Algebra, 6th edition Section 6.4 – Graphing Linear Equations Tom Atwater Draft 1 x + 3y = 1. READ Exercise 17 For the equation, complete the given ordered pair. 2x + 3y = 30; (0, ); (, 0) MATH STEPS: 2x + 3y = 30; (0, ) Substitute 0 for x and solve for y 2(0) + 3y = 30 3y = 30 y = 10 The ordered pair is (0, 10) is a solution of 2x + 3y = 30. Since, 2(0) + 3(10) = 30, the solution checks. 2x + 3y = 30; ( , 0) Substitute 0 for y and solve for x 2x + 3(0) = 30 2x = 30 x = 15 FULL SCREEN PRESENTER [PP 6] The ordered pair is (15, 0) is a solution of 2x + 3y = 30. Since, 2(15) + 3(0) = 30, the solution checks. SAY: In summary, here are the steps to finding a solution of an equation with two variables. READ To find a solution of an equation with two variables To find a solution of an equation with two variables 1. Choose a replacement for one variable. 2. Solve for the other variable. 3. Write the solution as an ordered pair. FULL SCREEN PRESENTER [PP 7] SAY: Here’s an example. READ Exercise 37 For the equation, find three solutions. Answers may vary. 2 Bittinger Pre-Algebra, 6th edition Section 6.4 – Graphing Linear Equations Tom Atwater Draft 1 3x + 2y = 12 MATH STEPS: 3x + 2y = 12 Substitute 1 for x and solve for y 3(1) + 2y = 12 2y = 9 y = 9/2 The ordered pair is (1, 9/2) is a solution of 3x + 2y = 12. In general, it is a good idea to use a replacement value of zero for both x and y. Let’s do that now. Substitute 0 for x and solve for y 3(0) + 2y = 12 2y = 12 y=6 The ordered pair is (0, 6) is a solution of 3x + 2y = 12. Substitute 0 for y and solve for x 3x + 2(0) = 12 3x = 12 x=4 The ordered pair is (4, 0) is a solution of 3x + 2y = 12. FULL SCREEN PRESENTER SAY: Next we consider graphing linear equations. Equations like those considered in Examples 1–3 are in the form WRITE ON THE BOARD Ax + By = C All equations that can be written this way are said to be linear because the solutions of each equation, when graphed, form a straight line. An equation Ax + By = C is called the standard form of a linear equation. When the line representing the 3 Bittinger Pre-Algebra, 6th edition Section 6.4 – Graphing Linear Equations Tom Atwater Draft 1 solutions is drawn, we say that we have graphed the equation. Since solutions of Ax + By = C are written in the form WRITE ON THE BOARD (x, y), FULL SCREEN PRESENTER [PP 8] we label the horizontal axis as the x-axis and the vertical axis as the y-axis. SAY: Let’s work two examples. READ Exercise 51 Graph the equation y = 2x – 1. MATH STEPS: First, we find some ordered pairs that are solutions. To find three ordered pairs, we can choose any three values for x and then calculate the corresponding values for y. We will organize these values in a table. One good choice is 0, and then a negative value and a positive value. x y=2x–1 (x, y) 0 –1 (0, –1) -2 –5 (–2, –5) 1 1 (1, 1) Plot these points and we draw the line, or graph, with a ruler and label it y = 2x – 1. [PP 9] READ Exercise 57 Graph the equation y = –x + 4. MATH STEPS: 4 Bittinger Pre-Algebra, 6th edition Section 6.4 – Graphing Linear Equations Tom Atwater Draft 1 Make a table. x y=–x+4 (x, y) 0 4 (0, 4) -3 7 (–3, 7) 2 2 (2, 2) Plot these points and we draw the line, or graph, with a ruler and label it y = –x + 4. FULL SCREEN PRESENTER [PP 10] SAY: Any equation in the form Ax + By = C is linear, provided A and B are not both zero. If A is 0 and B is nonzero, there is no x-term and the graph is a horizontal line. If B is 0 and A is nonzero, there is no y-term and the graph is a vertical line. Let’s work an example of each. READ Exercise 61 Graph y = 2. MATH STEPS: We regard y = 2 as 0 • x + y = 2. No matter what number we choose for x, we find that y must be 2 if the equation is to be solved. We can make a table to see this. x y=2 (x, y) 0 2 (0, 2) -3 2 (–3, 2) 2 2 (2, 2) When we plot these points and connect them, we obtain a horizontal line. Any ordered pair of the form (x, 2) is a solution, so the line is 2 units above the yaxis. 5 Bittinger Pre-Algebra, 6th edition Section 6.4 – Graphing Linear Equations Tom Atwater Draft 1 [PP 11] READ Exercise 65 Graph x = –3. MATH STEPS: No matter what number we choose for y, we find that x must be –3 if the equation is to be solved. Therefore, any ordered pair of the form (–3, y) is a solution, so the line is 3 units to the left of the x-axis. FULL SCREEN PRESENTER [PP 12] SAY: Let’s summarize what we just discovered. READ HORIZONTAL AND VERTICAL LINES HORIZONTAL AND VERTICAL LINES The graph of y = b is a horizontal line. The graph of x = a is a vertical line. FULL SCREEN PRESENTER CONCLUSION: Today we learned how to graph linear equations. Go back and review this section before beginning the exercise set. If you're having any trouble with the material see your instructor immediately. Good Luck. 6