1. ALGEBRA 1 Lesson 6-1 Warm-Up

2. ALGEBRA 1 “Solving Systems by Graphing” (6-1) What is a “system of linear equations”? What is the “solution of the system of linear equations”? How do you find the solution for a system of linear equations”? System of Linear Equations: two or more lines (linear equation) that are on the same graph Solution of the System of Linear Equations: a point that all of the lines in a system of equations have in common (in other words, all of the lines in the system of linear equations cross at that point) Method 1: Make a Graph – Graph all of the lines and find the point where all of the lines cross (the point of intersection) Example: Solve y = 2x – 3 and y = x – 1 Graph both equations on the same coordinate plane. Find the point of intersection. The lines intersect at (2,1), so (2,1) is the solution to the system.

3. ALGEBRA 1 “Solving Systems by Graphing” (6-1) (5-3)

4. ALGEBRA 1 Solve by graphing. Check your solution. y = 2x + 1 y = 3x – 1 Graph both equations on the same coordinate plane. y = 2x + 1The slope is 2. The y-intercept is 1. y = 3x – 1The slope is 3. The y-intercept is –1. Solving Systems by Graphing LESSON 6-1 Additional Examples

5. ALGEBRA 1 (continued) Find the point of intersection. The lines intersect at (2, 5), so (2, 5) is the solution of the system. Check: See if (2, 5) makes both equations true. y = 2x + 1 y = 3x – 1 5 2(2) + 1 Substitute (2, 5) for (x, y). 5 3(2) – 1 5 4 + 1 5 6 – 1 5 = 5 Solving Systems by Graphing LESSON 6-1 Additional Examples

6. ALGEBRA 1 Suppose you plan to start taking an aerobics class. Non-members pay $4 per class while members pay a $10 fee plus an additional $2 per class. Write a system of equations that models these plans. Define: Let = cost of one class Let = number of classes. Let = total cost of the classes. n C(n)C(n) Solving Systems by Graphing LESSON 6-1 Additional Examples C

7. ALGEBRA 1 The system is C(n) = 4n (or y = 4x) C(n) = 10 + 2n (or y = 2x + 10) Words: costismembershipplus cost of classes fee attended n C(n)C(n) Equation: member = 10 + 2 C(n)C(n) n non-member = 0 + 4 (continued) Solving Systems by Graphing LESSON 6-1 Additional Examples

8. ALGEBRA 1 The system below models the cost of taking an aerobics class as a function of the number of classes. Find the solution of the system by graphing. What does the solution mean in terms of the situation? C(a) = 2a + 10 The slope is 2. The intercept on the vertical axis is 10. C(a) = 2a + 10 Graph the equations. The lines intersect at (5, 20). Part 1: Find the solutions by graphing. C(a) = 4a The slope is 4. The intercept on the vertical axis is 0. C(a) = 4a Solving Systems by Graphing LESSON 6-1 Additional Examples

9. ALGEBRA 1 Part 2: Interpret the solution. (continued) Solving Systems by Graphing LESSON 6-1 Additional Examples The lines intersect at (5, 20) so the cost will be $20 after 5 classes.

10. ALGEBRA 1 “Solving Systems by Graphing” (6-1) When does a system of linear equations have no solutions? When does a system of linear equations have an infinite number of solutions? A system of linear equations has no solutions when the two (or more) lines are parallel to one another (in other words, there are no points of intersection). Example: Solve by graphing: y = -2x + 1 and y = -2x – 1 Graph both equations on the same coordinate plane. y = -2x + 1 The slope s -2. The y-intercept is 1 y = -2x - 1 The slope s -2. The y-intercept is -1. The lines are parallel, so there is no solution. A system of linear equations has an “infinite many solutions” when the graphs of the equation are the same line (in other words, every point on the line is a solution). Example: Solve by graphing:2x + 4y = 8 and y = - x + 2 Graph both equations on the same coordinate plane. 2x + 4y = 8 The y-intercept is 2 and the x-intercept is 4. y = - x + 2 The slope s -. The y-intercept is 2. The graphs are the same line, so there are an infinite number of solution 12 12 12 12 12 12

11. ALGEBRA 1 Solve by graphing. y = 3x + 2 y = 3x – 2 Graph both equations on the same coordinate plane. The lines are parallel. There is no solution. y = 3x + 2The slope is 3. The y-intercept is 2. y = 3x – 2 The slope is 3. The y-intercept is –2. Solving Systems by Graphing LESSON 6-1 Additional Examples

12. ALGEBRA 1 Solve by graphing.3x + 4y = 12 y = – x + 3 Graph both equations on the same coordinate plane. The graphs are the same line. The solutions are an infinite number of ordered pairs (x, y), such that y = – x + 3. 3434 3434 3x + 4y = 12The y-intercept is 3. The x-intercept is 4. y = – x + 3 The slope is –. The y-intercept is 3. 3434 3434 Solving Systems by Graphing LESSON 6-1 Additional Examples

13. ALGEBRA 1 1.y = –x – 22.y = –x + 3 3.y = 3x + 2 y = x + 3 y = 2x – 66x – 2y = –4 4. 2x – 3y = 9 5. –2x + 4y = 12 y = x – 5 – x + y = –3 2323 1212 (  3, 1) (3, 0)infinitely many solutions (6, 1)no solution Solve by graphing. Solving Systems by Graphing LESSON 6-1 Lesson Quiz