1 / 11

Solving Systems of Equations Algebraically: Substitution and Elimination Methods

This lesson focuses on solving systems of equations algebraically using substitution and elimination methods. Students will learn to find additive inverses, substitute variables, and solve for unknowns in various scenarios, including word problems. The lesson provides step-by-step guidance through examples, illustrating how to handle two-variable equations, and checking solutions for both accuracy and feasibility. By understanding these methods, students will enhance their problem-solving skills and grasp the concepts essential for advanced algebra studies.

mickey
Télécharger la présentation

Solving Systems of Equations Algebraically: Substitution and Elimination Methods

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. Section 3.2 Solving Systems Algebraically

  2. Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 (For help, go to Lesson 1-1 and 1-3.) Find the additive inverse of each term. 1. 4 2. –x3. 5x4. 8y Let x = 2y – 1. Substitute this expression for x in each equation. Solve for y. 5.x + 2y = 3 6.y – 2x = 8 7. 2y + 3x = –5 Check Skills You’ll Need 3-2

  3. 1. additive inverse of 4: –4 2. additive inverse of –x: x 3. additive inverse of 5x: –5x4. additive inverse of 8y: –8y 5.x + 2y = 3, with x = 2y – 1: (2y – 1) + 2y = 3 4y – 1 = 3 4y = 4 y = 1 7. 2y + 3x = –5, with x = 2y – 1: 2y + 3(2y – 1) = –5 2y + 6y – 3 = –5 8y – 3 = –5 8y = –2 y = – 6.y – 2x = 8, with x = 2y – 1: y – 2(2y – 1) = 8 y – 4y + 2 = 8 –3y + 2 = 8 –3y = 6 y = –2 1 4 Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 Solutions 3-2

  4. x + 3y = 12 –2x + 4y = 9 Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 Solve the system by substitution. Step 1:  Solve for one of the variables. Solving the first equation for x is the easiest. x + 3y = 12 x = –3y + 12 Step 2:  Substitute the expression for x into the other equation. Solve for y. –2x + 4y = 9 –2(–3y + 12) + 4y = 9 Substitute for x. 6y – 24 + 4y = 9 Distributive Property 6y + 4y = 33 y = 3.3 Step 3:  Substitute the value of y into either equation. Solve for x. x = –3(3.3) + 12 x = 2.1 Quick Check The solution is (2.1, 3.3). 3-2

  5. 2 p + s = 10.25 4 p + s = 18.75 Relate:  2 • price of a slice of pizza + price of a soda = $10.25 4 • price of a slice of pizza + price of a soda = $18.75 Define:  Let p = the price of a slice of pizza.     Let s = the price of a soda. Write: Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 At Renaldi’s Pizza, a soda and two slices of the pizza–of–the–day costs $10.25. A soda and four slices of the pizza–of–the–day costs $18.75. Find the cost of each item. 2p + s = 10.25 Solve for one of the variables. s = 10.25 – 2p 3-2

  6. Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 (continued) 4p + (10.25 – 2p) = 18.75 Substitute the expression for s into the other equation. Solve for p. p = 4.25 2(4.25) + s = 10.25 Substitute the value of p into one of the equations. Solve for s. s = 1.75 The price of a slice of pizza is $4.25, and the price of a soda is $1.75. Quick Check 3-2

  7. 3x + y = –9 –3x – 2y = 12 3x + y = –9 –3x – 2y = 12 Two terms are additive inverses, so add. –y = 3 Solve for y. Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 Use the elimination method to solve the system. y = –3 3x + y = –9 Choose one of the original equations. 3x + (–3) = –9 Substitute y.Solve for x. x = –2 The solution is (–2, –3). Quick Check 3-2

  8. 2m + 4n = –4 3m + 5n = –3 2m + 4n = –4 10m + 20n = –20 1 3m + 5n = –3–12m–20n = 12 2 –2m = –8 Add. Multiply by 5. 2 1 Multiply by –4. Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 Solve the system by elimination. To eliminate the n terms, make them additive inverses by multiplying. m = 4 Solve for m. 2m + 4n = –4 Choose one of the original equations. 2(4) + 4n = –4 Substitute for m. 8 + 4n = –4 4n = –12 Solve for n. n = –3 Quick Check The solution is (4, –3). 3-2

  9. –3x + 5y = 6 6x – 10y = 0 –6x + 10y = 12 6x – 10y = 0 Multiply the first line by 2 to make the x terms additive inverses. 0 = 12 Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 Solve each system by elimination. a. –3x + 5y = 6 6x – 10y = 0 Elimination gives an equation that is always false. The two equations in the system represent parallel lines. The system has no solution. 3-2 3-2

  10. –3x + 5y = 6 6x – 10y = –12 –6x + 10y = 12 6x + 10y = –12 Multiply the first line by 2 to make the x terms additive inverses. 0 = 0 Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 Quick Check (continued) b. –3x + 5y = 6 6x – 10y = –12 Elimination gives an equation that is always true. The two equations in the system represent the same line. The system has an infinite number of solutions. 3-2

  11. 1. Solve by substitution. 2. A bookstore took in $167 on the sale of 5 copies of a new cookbook and 3 copies of a new novel. The next day it took in $89 on the sales of 3 copies of the cookbook and 1 copy of the novel. What was the price of each book? Solve each system. 3.4. 5. –2x + 5y = –2 x – 3y = 3 10x + 6y = 0 –7x + 2y = 31 7x + 5y = 18 –7x – 9y = 4 –3x + y = 6 6x – 2y = 25 Solving Systems Algebraically ALGEBRA 2 LESSON 3-2 (–9, –4) cookbook: $25; novel: $14 (–3, 5) (6.5, –5.5) no solutions 3-2

More Related