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Warm Up ( 10mins)

Warm Up ( 10mins). Algebra 2 Textbook Page 123 Check Skills You’ll Need Exercises # 2, 4, 5-7. 1. additive inverse of 4: –4 2. additive inverse of – x : x 3. additive inverse of 5 x : –5 x 4. additive inverse of 8 y : –8 y 5. x + 2 y = 3, with x = 2 y – 1:

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Warm Up ( 10mins)

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  1. Warm Up (10mins) Algebra 2 Textbook Page 123 Check Skills You’ll Need Exercises # 2, 4, 5-7

  2. 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 Solutions

  3. LINEAR SYSTEMS Lesson 3.2: SOLVING SYSTEMS ALGEBRAICALLY

  4. Objective • Students will use substitution and elimination in order to solve problems using systems of linear equations.

  5. INTRODUCTORY AND DEVELOPMENTAL ACTIVITIES Vocabulary: Find the following terms and pages 125 and write down the definitions in your notebook: •Equivalent systems

  6. To solve a system of equations by substitution Step 1: Solve for one of the variables. Choose the easiest one. Step 2: Substitute the expression of the variable that you choose into the other equation. Then solve the equation. Step 3: Substitute the value of the variable found into either equation . Then solve the equation.

  7. x + 3y = 12 –2x + 4y = 9 Solving Systems Algebraically 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 = 9Substitute 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 The solution is (2.1, 3.3).

  8. 2p+ s = 10.25 4p+ 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.     Lets= the price of a soda. Write: Solving Systems Algebraically 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

  9. Solving Systems Algebraically (continued) 4p + (10.25 – 2p) = 18.75Substitute 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.

  10. Pages 123-124 Complete check understanding 1-2, # 5, 14. Work the steps thoroughly for a better comprehension. Post # 14 on discussion board, and respond to at least two classmates.

  11. To solve system of equations by elimination Step 1: Decide on which variable to eliminate Step 2: Multiply equation 1 by the coefficient of the variable that you want to eliminate in equation 2 and equation 2 by the coefficient of the variable that you want to eliminate in equation 1 Step 3: Add or subtract the two equations Step 4:Solve

  12. 3x + y = –9 –3x – 2y = 12 3x + y = –9 –3x – 2y = 12Two terms are additive inverses, so add. –y = 3Solve for y. Solving Systems Algebraically 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).

  13. 2m + 4n = –4 3m + 5n = –3 2m + 4n = –4 10m + 20n = –20Multiply by 5. 1 3m + 5n = –3–12m–20n = 12Multiply by –4. 2 –2m = –8 Add. 1 2 Solving Systems Algebraically 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 The solution is (4, –3).

  14. –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 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.

  15. –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 {(x, y)| y = x + } 3 5 6 5 Solving Systems Algebraically Solve each system by elimination. 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:

  16. Pages 124-125 Complete check understanding 3-5, # 32, 42. Work the steps thoroughly for a better comprehension. Post # 42 on the discussion board, respond to at least two classmates.

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