Solving special systems
- 1. Holt McDougal Algebra 1 5-4 Solving Special Systems5-4 Solving Special Systems Holt Algebra 1 Warm UpWarm Up Lesson PresentationLesson Presentation Lesson QuizLesson Quiz Holt McDougal Algebra 1
- 2. Holt McDougal Algebra 1 5-4 Solving Special Systems Solve special systems of linear equations in two variables. Classify systems of linear equations and determine the number of solutions. Objectives
- 3. Holt McDougal Algebra 1 5-4 Solving Special Systems Example 1: Systems with No Solution Method 1 Compare slopes and y-intercepts. y = x – 4 y = 1x – 4 Write both equations in slope- intercept form. –x + y = 3 y = 1x + 3 Show that has no solution. y = x – 4 –x + y = 3 The lines are parallel because they have the same slope and different y-intercepts. This system has no solution.
- 4. Holt McDougal Algebra 1 5-4 Solving Special Systems Example 1 Continued Method 2 Solve the system algebraically. Use the substitution method because the first equation is solved for y. –x + (x – 4) = 3 Substitute x – 4 for y in the second equation, and solve. –4 = 3 False. This system has no solution. Show that has no solution. y = x – 4 –x + y = 3
- 5. Holt McDougal Algebra 1 5-4 Solving Special Systems Example 1 Continued Check Graph the system. The lines appear are parallel. – x + y = 3 y = x – 4 Show that has no solution. y = x – 4 –x + y = 3
- 6. Holt McDougal Algebra 1 5-4 Solving Special Systems Check It Out! Example 1 Method 1 Compare slopes and y-intercepts. Show that has no solution. y = –2x + 5 2x + y = 1 y = –2x + 5 y = –2x + 5 2x + y = 1 y = –2x + 1 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. Holt McDougal Algebra 1 5-4 Solving Special Systems Method 2 Solve the system algebraically. Use the substitution method because the first equation is solved for y. 2x + (–2x + 5) = 1 Substitute –2x + 5 for y in the second equation, and solve. False. This system has no solution. 5 = 1 Check It Out! Example 1 Continued Show that has no solution. y = –2x + 5 2x + y = 1
- 8. Holt McDougal Algebra 1 5-4 Solving Special Systems Check Graph the system. The lines are parallel. y = – 2x + 1 y = –2x + 5 Check It Out! Example 1 Continued Show that has no solution. y = –2x + 5 2x + y = 1
- 9. Holt McDougal Algebra 1 5-4 Solving Special Systems Show that has infinitely many solutions. y = 3x + 2 3x – y + 2= 0 Example 2A: Systems with Infinitely Many Solutions Method 1 Compare slopes and y-intercepts. y = 3x + 2 y = 3x + 2 Write both equations in slope- intercept form. The lines have the same slope and the same y-intercept. 3x – y + 2= 0 y = 3x + 2 If this system were graphed, the graphs would be the same line. There are infinitely many solutions.
- 10. Holt McDougal Algebra 1 5-4 Solving Special Systems Method 2 Solve the system algebraically. Use the elimination method. y = 3x + 2 y − 3x = 2 3x − y + 2= 0 −y + 3x = −2 Write equations to line up like terms. Add the equations. True. The equation is an identity. 0 = 0 There are infinitely many solutions. Example 2A Continued Show that has infinitely many solutions. y = 3x + 2 3x – y + 2= 0
- 11. Holt McDougal Algebra 1 5-4 Solving Special Systems Check It Out! Example 2 Show that has infinitely many solutions. y = x – 3 x – y – 3 = 0 Method 1 Compare slopes and y-intercepts. y = x – 3 y = 1x – 3 Write both equations in slope- intercept form. The lines have the same slope and the same y-intercept. x – y – 3 = 0 y = 1x – 3 If this system were graphed, the graphs would be the same line. There are infinitely many solutions.
- 12. Holt McDougal Algebra 1 5-4 Solving Special Systems 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. y = x – 3 y = x – 3 x – y – 3 = 0 –y = –x + 3 0 = 0 There are infinitely many solutions. Check It Out! Example 2 Continued Show that has infinitely many solutions. y = x – 3 x – y – 3 = 0
- 13. Holt McDougal Algebra 1 5-4 Solving Special Systems Example 3B: Classifying Systems of Linear equations Solve x + y = 5 4 + y = –x Classify the system. Give the number of solutions. x + y = 5 y = –1x + 5 4 + y = –x y = –1x – 4 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. Holt McDougal Algebra 1 5-4 Solving Special Systems Example 3C: Classifying Systems of Linear equations Classify the system. Give the number of solutions. Solve y = 4(x + 1) y – 3 = x y = 4(x + 1) y = 4x + 4 y – 3 = x y = 1x + 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.
- 15. Holt McDougal Algebra 1 5-4 Solving Special Systems Check It Out! Example 3a Classify the system. Give the number of solutions. Solve x + 2y = –4 –2(y + 2) = x Write both equations in slope-intercept form. y = x – 2x + 2y = –4 –2(y + 2) = x y = x – 2 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. Holt McDougal Algebra 1 5-4 Solving Special Systems Check It Out! Example 3b Classify the system. Give the number of solutions. Solve y = –2(x – 1) y = –x + 3 y = –2(x – 1) y = –2x + 2 y = –x + 3 y = –1x + 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. Holt McDougal Algebra 1 5-4 Solving Special Systems Check It Out! Example 3c Classify the system. Give the number of solutions. Solve 2x – 3y = 6 y = x y = x y = x 2x – 3y = 6 y = x – 2 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.