1 / 19

3.1

3.1. Solve a system graphically. – 1. ANSWER. C = 18 x. ANSWER. Lesson 1.1 , For use with pages 153-158. 3. Find the x - intercept of the graph of y = . . | x + 1| . 4. Express the cost C of x ball game tickets at a price of $18 per ticket.

ryu
Download Presentation

3.1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 3.1 Solve a system graphically

  2. – 1 ANSWER C = 18 x ANSWER Lesson 1.1, For use with pages 153-158 3.Find thex-intercept of the graph ofy = . | x + 1| 4.Express the costC ofx ball game tickets at a price of $18 per ticket.

  3. Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (3, – 4). You can check this algebraically as follows. EXAMPLE 1 Solve a system graphically Graph the linear system and estimate the solution. Then check the solution algebraically. 4x + y = 8 Equation 1 2x – 3y = 18 Equation 2 SOLUTION

  4. 4(3) +(–4) 8 8 = 8 2(3) – 3(– 4) 18 6 + 12 18 12 – 4 8 18 = 18 ? ? ? ? = = = = EXAMPLE 1 Solve a system graphically Equation 2 Equation 1 4x+ y= 8 2x– 3y= 18 The solution is (3, – 4).

  5. 1. 3x + 2y = – 4 x + 3y = 1 Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–2, 1). You can check this algebraically as follows. for Example 1 GUIDED PRACTICE Graph the linear system and estimate the solution. Then check the solution algebraically. 3x + 2y = – 4 Equation 1 x + 3y = 1 Equation 2 SOLUTION

  6. (–2 ) + 3(1) 1 ? = 3(–2) +2(1) –4 –2 + 3 1 –4 = –4 –6 + 2 –4 1 = 1 ? ? ? = = = for Example 1 GUIDED PRACTICE Equation 2 Equation 1 3x+ 2y= –4 x+ 3y= 1 The solution is (–2, 1).

  7. 2. 4x – 5y = – 10 2x – 7y = 4 Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (–5, –2). You can check this algebraically as follows. for Example 1 GUIDED PRACTICE Graph the linear system and estimate the solution. Then check the solution algebraically. 4x – 5y = – 10 Equation 1 2x – 7y = 4 Equation 2 SOLUTION

  8. 2(–5)– 7(–2) 4 4(–5) –5(–2 ) – 10 ? = –20 + 10 –10 –10 + 14 4 –10 = –10 4 = 4 ? ? ? = = = for Example 1 GUIDED PRACTICE Equation 2 Equation 1 4x – 5y = –10 2x–7y= 4 The solution is (–5, –2).

  9. 3. 8x – y = 8 3x + 2y = – 16 Begin by graphing both equations, as shown at the right. From the graph, the lines appear to intersect at (0, –8). You can check this algebraically as follows. for Example 1 GUIDED PRACTICE Graph the linear system and estimate the solution. Then check the solution algebraically. 8x – y = 8 Equation 1 3x + 2y = – 16 Equation 2 SOLUTION

  10. 3(0) + 2(–8) –16 8(0) –(–8 ) 8 ? = 0 – 16 –16 –16 = –16 8 = 8 0 + 8 8 ? ? ? = = = for Example 1 GUIDED PRACTICE Equation 2 Equation 1 8x– y = 8 3x+ 2y= –16 The solution is (0, –8).

  11. The graphs of the equations are the same line. So, each point on the line is a solution, and the system has infinitely many solutions. Therefore, the system is consistent and dependent. EXAMPLE 2 Solve a system with many solutions Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent. 4x – 3y = 8 Equation 1 8x – 6y = 16 Equation 2 SOLUTION

  12. The graphs of the equations are two parallel lines.Because the two lines have no point of intersection, the system has no solution. Therefore, the system is inconsistent. EXAMPLE 3 Solve a system with no solution Solve the system. Then classify the system as consistent and independent,consistent and dependent, or inconsistent. Equation 1 2x + y = 4 Equation 2 2x + y = 1 SOLUTION

  13. y x 1 30 + = EXAMPLE 4 Standardized Test Practice SOLUTION Equation 1 (Option A)

  14. x 2.5 y = EXAMPLE 4 Standardized Test Practice Equation 2 (Option B) To solve the system, graph the equations y = x+ 30 and y = 2.5x, as shown at the right.

  15. 50= 20+ 30 50= 2.5(20) The total costs are equal after 20 rides. ANSWER The correct answer is B. EXAMPLE 4 Standardized Test Practice Notice that you need to graph the equations only in the first quadrant because only nonnegative values of xand ymake sense in this situation. The lines appear to intersect at about the point (20, 50). You can check this algebraically as follows. Equation 1 checks. Equation 2 checks.

  16. Infinitely many solutions; consistent and dependent ANSWER for Examples 2,3, and 4 GUIDED PRACTICE Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent. 4. 2x + 5y = 6 4x + 10y = 12

  17. 5. 3x – 2y = 10 3x – 2y = 2 no solution; inconsistent ANSWER for Examples 2,3, and 4 GUIDED PRACTICE Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.

  18. ANSWER (–1, 3); consistent and independent – 2x + y = 5 6. y = – x + 2 for Examples 2,3, and 4 GUIDED PRACTICE Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.

  19. ANSWER The number of rides increases to 24. 7. WHAT IF? In Example 4, suppose the cost of the monthly pass is increased to $36. How does this affect the solution? for Examples 2,3, and 4 GUIDED PRACTICE Solve the system. Then classify the system as consistent and independent, consistent and dependent, or inconsistent.

More Related