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Solving Systems Using Elimination (For help, go to Lesson 7-2.) Solve each system using substitution. 1.y = 4x – 32.y + 5x = 4 3. y = –2x + 2 y = 2x + 13y = 7x – 20 3x – 17 = 2y ALGEBRA 1 LESSON 9-5 9-5

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Solving Systems Using Elimination 1. y = 4x – 3 y = 2x + 13 Substitute 4x – 3 for y in the second equation. y = 2x + 13 4x – 3 = 2x + 13 4x – 2x – 3 = 2x – 2x + 13 2x – 3 = 13 2x = 16 x = 8 y = 4x – 3 = 4(8) – 3 = 32 – 3 = 29 Since x = 8 and y = 29, the solution is (8, 29). ALGEBRA 1 LESSON 9-5 Solutions 9-5

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Solving Systems Using Elimination 2. y + 5x = 4 y = 7x – 20 Substitute 7x – 20 for y in the first equation. y + 5x = 4 7x – 20 + 5x = 4 12x – 20 = 4 12x = 24 x = 2 y = 7x – 20 = 7(2) – 20 = 14 – 20 = –6 Since x = 2 and y = –6, the solution is (2, –6). ALGEBRA 1 LESSON 9-5 Solutions (continued) 9-5

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Solving Systems Using Elimination 3. y = –2x + 2 3x – 17 = 2y Substitute –2x + 2 for y in the second equation. 3x – 17 = 2y 3x – 17 = 2(–2x + 2) 3x – 17 = –4x + 4 7x – 17 = 4 7x = 21 x = 3 y = –2x + 2 = –2(3) + 2 = –6 + 2 –4 Since x = 3 and y = –4, the solution is (3, –4). ALGEBRA 1 LESSON 9-5 Solutions (continued) 9-5

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Solving Systems Using Elimination Solve by elimination.3x + 6y = –6 –5x – 2y = –14 Step 1:Eliminate one variable. Start with the given system. 3x + 6y = –6 –5x – 2y = –14 Add the equations to eliminate y. 3x + 6y = –6 –15x – 6y = –42 –12x – 0 = –48 To prepare to eliminate y, multiply the second equation by 3. 3x + 6y = –6 3(–5x – 2y = –14) Step 2:Solve for x. –12x = 48 x = 4 ALGEBRA 1 LESSON 9-5 9-5

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Solving Systems Using Elimination (continued) Step 3:Solve for the eliminated variable using either of the original equations. 3x + 6y = –6Choose the first equation. 3(4) + 6y = –6Substitute 4 for x. 12 + 6y = –6Solve for y. 6y = –18 y = –3 The solution is (4, –3). ALGEBRA 1 LESSON 9-5 9-5

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Define:Let p = number of cans of popcorn sold. Let n = number of cans of nuts sold. Relate:total number of canstotal amount of sales Write: p + n = 2405 p + 8 n = 1614 Solving Systems Using Elimination Suppose the band sells cans of popcorn for $5 per can and cans of mixed nuts for $8 per can. The band sells a total of 240 cans and receives a total of $1614. Find the number of cans of popcorn and the number of cans of mixed nuts sold. ALGEBRA 1 LESSON 9-5 9-5

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Solving Systems Using Elimination (continued) Step 1:Eliminate one variable. Subtract the equations to eliminate p. 5p + 5n = 1200 5p + 8n = 1614 0 – 3n = –414 Step 2:Solve for n. –3n = –414 n = 138 Start with the given system. p + n = 240 5p + 8n = 1614 To prepare to eliminate p, multiply the first equation by 5. 5(p + n = 240) 5p + 8n = 1614 ALGEBRA 1 LESSON 9-5 9-5

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Solving Systems Using Elimination (continued) Step 3:Solve for the eliminated variable using either of the original equations. p + n = 240Choose the first equation. p + 138 = 240Substitute 138 for n. p = 102Solve for p. The band sold 102 cans of popcorn and 138 cans of mixed nuts. ALGEBRA 1 LESSON 9-5 9-5

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Solving Systems Using Elimination Solve by elimination.3x + 5y = 10 5x + 7y = 10 Step 1:Eliminate one variable. Subtract the equations to eliminate x. 15x + 25y = 50 15x + 21y = 30 0 + 4y = 20 Start with the given system. 3x + 5y = 10 5x + 7y = 10 To prepare to eliminate x, multiply one equation by 5 and the other equation by 3. 5(3x + 5y = 10) 3(5x + 7y = 10) Step 2:Solve for y. 4y = 20 y = 5 ALGEBRA 1 LESSON 9-5 9-5

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Solving Systems Using Elimination (continued) Step 3:Solve for the eliminated variable x using either of the original equations. 3x + 5y = 10Use the first equation. 3x + 5(5) = 10Substitute 5 for y. 3x + 25 = 10 3x = –15 x = –5 The solution is (–5, 5). ALGEBRA 1 LESSON 9-5 9-5

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Solving Systems Using Elimination Solve using elimination. 1. –6x + 5y = 4 2. 7p + 5q = 2 3x + 4y = 11 8p – 9q = 17 (1, 2) (1, 1) ALGEBRA 1 LESSON 9-5 9-5

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