Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our User Agreement and Privacy Policy.

Slideshare uses cookies to improve functionality and performance, and to provide you with relevant advertising. If you continue browsing the site, you agree to the use of cookies on this website. See our Privacy Policy and User Agreement for details.

Successfully reported this slideshow.

Like this presentation? Why not share!

- 4 1 solving linear systems by graphing by hisema01 645 views
- Mathematics 9 Radical Expressions (2) by jmpalero 714 views
- Taxonomy and Bacteria PowerPoint Re... by www.sciencepowerp... 783 views
- Algebra 2. 9.16 Quadratics 2 by dmatkeson21 2146 views
- Algebra 1. 9.12 Lesson. Proportions by dmatkeson21 2139 views
- Ca mod06 les01 by Robert Hill 407 views

No Downloads

Total views

4,755

On SlideShare

0

From Embeds

0

Number of Embeds

22

Shares

0

Downloads

76

Comments

2

Likes

2

No embeds

No notes for slide

- 1. 6-1 Solving Systems by Graphing Holt Algebra 1 Warm Up Lesson Presentation Lesson Quiz
- 2. Warm Up Evaluate each expression for x = 1 and y =–3. 1. x – 4 y 2. –2 x + y Write each expression in slope-intercept form. 3. y – x = 1 4. 2 x + 3 y = 6 5. 0 = 5 y + 5 x 13 – 5 y = x + 1 y = x + 2 y = – x
- 3. Identify solutions of linear equations in two variables. Solve systems of linear equations in two variables by graphing. Objectives
- 4. systems of linear equations solution of a system of linear equations Vocabulary
- 5. A system of linear equations is a set of two or more linear equations containing two or more variables. A solution of a system of linear equations with two variables is an ordered pair that satisfies each equation in the system. So, if an ordered pair is a solution, it will make both equations true.
- 6. Tell whether the ordered pair is a solution of the given system. Example 1A: Identifying Systems of Solutions (5, 2); The ordered pair (5, 2) makes both equations true. (5, 2) is the solution of the system. Substitute 5 for x and 2 for y in each equation in the system. 3 x – y = 13 2 – 2 0 0 0 0 3 (5) – 2 13 15 – 2 13 13 13 3 x – y 13
- 7. If an ordered pair does not satisfy the first equation in the system, there is no reason to check the other equations. Helpful Hint
- 8. Example 1B: Identifying Systems of Solutions Tell whether the ordered pair is a solution of the given system. (–2, 2); x + 3 y = 4 – x + y = 2 Substitute –2 for x and 2 for y in each equation in the system. The ordered pair (–2, 2) makes one equation true but not the other. (–2, 2) is not a solution of the system. – 2 + 3 (2) 4 x + 3 y = 4 – 2 + 6 4 4 4 – x + y = 2 – (–2) + 2 2 4 2
- 9. Check It Out! Example 1a Tell whether the ordered pair is a solution of the given system. The ordered pair (1, 3) makes both equations true. Substitute 1 for x and 3 for y in each equation in the system. (1, 3) is the solution of the system. (1, 3); 2 x + y = 5 – 2 x + y = 1 2 x + y = 5 2 (1) + 3 5 2 + 3 5 5 5 – 2 x + y = 1 – 2 (1) + 3 1 – 2 + 3 1 1 1
- 10. Check It Out! Example 1b Tell whether the ordered pair is a solution of the given system. (2, –1); x – 2 y = 4 3 x + y = 6 The ordered pair (2, –1) makes one equation true, but not the other. Substitute 2 for x and –1 for y in each equation in the system. (2, –1) is not a solution of the system. 3 x + y = 6 3 (2) + (–1) 6 6 – 1 6 5 6 x – 2 y = 4 2 – 2 (–1) 4 2 + 2 4 4 4
- 11. All solutions of a linear equation are on its graph. To find a solution of a system of linear equations, you need a point that each line has in common. In other words, you need their point of intersection. The point (2, 3) is where the two lines intersect and is a solution of both equations, so (2, 3) is the solution of the systems. y = 2 x – 1 y = – x + 5
- 12. Sometimes it is difficult to tell exactly where the lines cross when you solve by graphing. It is good to confirm your answer by substituting it into both equations. Helpful Hint
- 13. Solve the system by graphing. Check your answer. Example 2A: Solving a System Equations by Graphing y = x y = –2 x – 3 Graph the system. The solution appears to be at (–1, –1). (–1, –1) is the solution of the system. y = x y = –2 x – 3 • (–1, –1) Check Substitute (–1, –1) into the system. y = x (–1) (–1) – 1 –1 y = –2 x – 3 ( – 1) –2 ( – 1) –3 – 1 2 – 3 – 1 – 1
- 14. Solve the system by graphing. Check your answer. Example 2B: Solving a System Equations by Graphing y = x – 6 Rewrite the second equation in slope-intercept form. Graph using a calculator and then use the intercept command. y + x = –1 y = x – 6 y + x = –1 − x − x y =
- 15. Solve the system by graphing. Check your answer. Example 2B Continued Check Substitute into the system. y = x – 6 The solution is . + – 1 – 1 – 1 – 1 – 1 y = x – 6 – 6
- 16. Solve the system by graphing. Check your answer. Check It Out! Example 2a y = –2 x – 1 y = x + 5 Graph the system. The solution appears to be (–2, 3). Check Substitute (–2, 3) into the system. (–2, 3) is the solution of the system. y = x + 5 3 – 2 + 5 3 3 y = –2 x – 1 3 –2 ( – 2) – 1 3 4 – 1 3 3 y = x + 5 y = –2 x – 1
- 17. Solve the system by graphing. Check your answer. Check It Out! Example 2b 2 x + y = 4 Rewrite the second equation in slope-intercept form. Graph using a calculator and then use the intercept command. 2 x + y = 4 – 2 x – 2 x y = –2 x + 4 2 x + y = 4
- 18. Solve the system by graphing. Check your answer. Check It Out! Example 2b Continued 2 x + y = 4 The solution is (3, –2). Check Substitute (3, –2) into the system. 2 x + y = 4 2 (3) + (–2) 4 6 – 2 4 4 4 2 x + y = 4 – 2 (3) – 3 – 2 1 – 3 – 2 –2
- 19. Lesson Quiz: Part I Tell whether the ordered pair is a solution of the given system. 1. (–3, 1); 2. (2, –4); yes no
- 20. Lesson Quiz: Part II Solve the system by graphing. 3. (2, 5) y + 2 x = 9 y = 4 x – 3

No public clipboards found for this slide

x-2y=-5