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Section 3.2

Section 3.2. Solving Systems Algebraically. 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. – x 3. 5 x 4. 8 y Let x = 2 y – 1. Substitute this expression for x in each equation.

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Section 3.2

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  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

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