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5-4. Solving Special Systems. Warm Up. Lesson Presentation. Lesson Quiz. Holt McDougal Algebra 1. Holt Algebra 1. Objectives. Solve special systems of linear equations in two variables. Classify systems of linear equations and determine the number of solutions. y = x – 4.

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5-4

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  1. 5-4 Solving Special Systems Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 1 Holt Algebra 1

  2. Objectives Solve special systems of linear equations in two variables. Classify systems of linear equations and determine the number of solutions.

  3. y = x – 4 Show that has no solution. –x + y = 3 y = x – 4 y = 1x –4 –x + y = 3 y = 1x + 3 Example 1: Systems with No Solution Method 1 Compare slopes and y-intercepts. Write both equations in slope-intercept form. The lines are parallel because they have the same slope and different y-intercepts. This system has no solution.

  4. y = x – 4 Show that has no solution. –x + y = 3  –4 = 3 Example 1 Continued Method 2 Solve the system algebraically. Use the substitution method because the first equation is solved for y. Substitute x – 4 for y in the second equation, and solve. –x + (x – 4) = 3 False. This system has no solution.

  5. y = x – 4 Show that has no solution. –x + y = 3 Example 1 Continued Check Graph the system. –x + y = 3 The lines appear are parallel. y = x– 4

  6. y =–2x + 5 Show that has no solution. 2x + y =1 y = –2x + 5 y = –2x + 5 2x + y = 1 y = –2x + 1 Check It Out! Example 1 Method 1 Compare slopes and y-intercepts. Write both equations in slope-intercept form. The lines are parallel because they have the same slope and different y-intercepts. This system has no solution.

  7. y =–2x + 5 Show that has no solution. 2x + y =1  5 = 1 Check It Out! Example 1 Continued Method 2 Solve the system algebraically. Use the substitution method because the first equation is solved for y. Substitute –2x + 5 for y in the second equation, and solve. 2x +(–2x + 5) = 1 False. This system has no solution.

  8. y =–2x + 5 Show that has no solution. 2x + y =1 Check It Out! Example 1 Continued Check Graph the system. y = –2x + 5 y = – 2x + 1 The lines are parallel.

  9. y =3x + 2 Show that has infinitely many solutions. 3x – y +2= 0 y = 3x + 2 y = 3x + 2 3x – y + 2= 0 y = 3x + 2 Example 2A: Systems with Infinitely Many Solutions Method 1 Compare slopes and y-intercepts. Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept. If this system were graphed, the graphs would be the same line. There are infinitely many solutions.

  10. y =3x + 2 Show that has infinitely many solutions. 3x – y +2= 0 y = 3x + 2 y − 3x = 2 3x − y + 2= 0 −y + 3x = −2  0 = 0 Example 2A Continued Method 2 Solve the system algebraically. Use the elimination method. Write equations to line up like terms. Add the equations. True. The equation is an identity. There are infinitely many solutions.

  11. y = x – 3 Show that has infinitely many solutions. x – y –3 = 0 y = x – 3 y = 1x –3 x – y –3 = 0 y = 1x –3 Check It Out! Example 2 Method 1 Compare slopes and y-intercepts. Write both equations in slope-intercept form. The lines have the same slope and the same y-intercept. If this system were graphed, the graphs would be the same line. There are infinitely many solutions.

  12. y = x – 3 Show that has infinitely many solutions. x – y –3 = 0 y = x – 3 y = x –3 x – y –3 = 0 –y = –x + 3  0 = 0 Check It Out! Example 2 Continued Method 2 Solve the system algebraically. Use the elimination method. Write equations to line up like terms. Add the equations. True. The equation is an identity. There are infinitely many solutions.

  13. x + y = 5 y = –1x + 5 4 + y = –x y = –1x – 4 Example 3B: Classifying Systems of Linear equations Classify the system. Give the number of solutions. x + y = 5 Solve 4+ y = –x Write both equations in slope-intercept form. The lines have the same slope and different y-intercepts. They are parallel. The system is inconsistent. It has no solutions.

  14. y = 4(x + 1) y = 4x + 4 y –3 = x y = 1x + 3 Example 3C: Classifying Systems of Linear equations Classify the system. Give the number of solutions. y = 4(x + 1) Solve y – 3 = x Write both equations in slope-intercept form. The lines have different slopes. They intersect. The system is consistent and independent. It has one solution.

  15. x + 2y = –4 y = x –2 –2(y + 2) = x y = x –2 Check It Out! Example 3a Classify the system. Give the number of solutions. x + 2y = –4 Solve –2(y + 2) = x Write both equations in slope-intercept form. The lines have the same slope and the same y-intercepts. They are the same. The system is consistent and dependent. It has infinitely many solutions.

  16. y = –2(x – 1) y = –2x + 2 y = –x + 3 y = –1x + 3 Check It Out! Example 3b Classify the system. Give the number of solutions. y = –2(x – 1) Solve y = –x + 3 Write both equations in slope-intercept form. The lines have different slopes. They intersect. The system is consistent and independent. It has one solution.

  17. 2x – 3y = 6 y = x –2 y = x y = x Check It Out! Example 3c Classify the system. Give the number of solutions. 2x – 3y = 6 Solve y = x Write both equations in slope-intercept form. The lines have the same slope and different y-intercepts. They are parallel. The system is inconsistent. It has no solutions.

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