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# Solving Systems Using Elimination

2 x + 3 y = 11 – 2 x + 9 y =1 0 + 12 y = 12 Addition Property of Equality y = 1 Solve for y. Solving Systems Using Elimination. Lesson 7-3. Additional Examples. Solve by elimination. 2 x + 3 y = 11 – 2 x + 9 y = 1.

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## Solving Systems Using Elimination

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1. 2x+ 3y = 11 –2x+ 9y =1 0+ 12y = 12 Addition Property of Equality y = 1 Solve for y. Solving Systems Using Elimination Lesson 7-3 Additional Examples Solve by elimination. 2x+ 3y = 11 –2x+ 9y = 1 Step 1: Eliminate x because the sum of the coefficients is 0. Step 2: Solve for the eliminated variable x using either original equation. 2x+ 3y = 11 Choose the first equation. 2x+ 3(1) = 11 Substitute 1 for y. 2x+ 3 = 11 Solve for x. 2x= 8 x= 4

2. Check: See if (4, 1) makes true the equation not used in Step 2. –2(4) + 9(1) 1 Substitute 4 for x and 1 for y into the second equation. –8 + 9 1 1 = 1 Solving Systems Using Elimination Lesson 7-3 Additional Examples (continued) Since x= 4 and y = 1, the solution is (4, 1).

3. Define: Let a= number of adults Let s = number of students Relate: total number at the game total amount collected Write: a + s = 1139 5 a + s = 3067 Solving Systems Using Elimination Lesson 7-3 Additional Examples On a special day, tickets for a minor league baseball game were \$5 for adults and \$1 for students. The attendance that day was 1139, and \$3067 was collected. Write and solve a system of equations to find the number of adults and the number of students that attended the game. Solve by elimination.

4. Solving Systems Using Elimination Lesson 7-3 Additional Examples (continued) Step 1: Eliminate one variable. a+ s = 1139 5a+ s = 3067 –4a+ 0 = –1928 Subtraction Property of Equality a = 482 Solve for a. Step 2: Solve for the eliminated variable using either of the original equations. a+ s = 1139 Choose the first equation. 482+ s = 1139 Substitute 482 for a. s = 657 Solve for s. There were 482 adults and 657 students at the game. Check: Is the solution reasonable? The total number at the game was 482 + 657, or 1139. The money collected was \$5(482), or \$2410, plus \$1(657), or \$657, which is \$3067. The solution is correct.

5. Start with the given system. 3x+ 6y = –6 –5x– 2y = –14 To prepare to eliminate y, multiply the second equation by 3. 3x+ 6y = –6 3(–5x– 2y = –14) Add the equations to eliminate y. 3x+ 6y = –6 –15x– 6y = –42 –12x– 0 = –48 Solving Systems Using Elimination Lesson 7-3 Additional Examples Solve by elimination. 3x+ 6y = –6 –5x– 2y = –14 Step 1: Eliminate one variable. Step 2: Solve for x. –12x= 48 x= 4

6. Solving Systems Using Elimination Lesson 7-3 Additional Examples (continued) Step 3: Solve for the eliminated variable using either of the original equations. 3x+ 6y = –6 Choose the first equation. 3(4) + 6y = –6 Substitute 4 for x. 12 + 6y = –6 Solve for y. 6y = –18 y = –3 The solution is (4, –3).

7. Define: Let p= number of cans of popcorn sold. Let n = number of cans of nuts sold. Relate: total number of cans total amount of sales Write: p + n = 240 5 p + 8 n = 1614 Solving Systems Using Elimination Lesson 7-3 Additional Examples 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.

8. 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 Subtract the equations to eliminate p. 5p+ 5n = 1200 5p+ 8n = 1614 0 – 3n = –414 Solving Systems Using Elimination Lesson 7-3 Additional Examples (continued) Step 1: Eliminate one variable. Step 2: Solve for n. –3n = –414 n = 138

9. Solving Systems Using Elimination Lesson 7-3 Additional Examples (continued) Step 3: Solve for the eliminated variable using either of the original equations. p+ n = 240 Choose the first equation. p+ 138 = 240 Substitute 138 for n. p = 102 Solve for p. The band sold 102 cans of popcorn and 138 cans of mixed nuts.

10. 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) Subtract the equations to eliminate x. 15x+ 25y = 50 15x+ 21y = 30 0 + 4y = 20 Solving Systems Using Elimination Lesson 7-3 Additional Examples Solve by elimination. 3x+ 5y = 10 5x + 7y = 10 Step 1: Eliminate one variable. Step 2: Solve for y. 4y = 20 y = 5

11. Solving Systems Using Elimination Lesson 7-3 Additional Examples (continued) Step 3: Solve for the eliminated variable x using either of the original equations. 3x+ 5y = 10 Use the first equation. 3x+ 5(5) = 10 Substitute 5 for y. 3x+ 25 = 10 3x = –15 x = –5 The solution is (–5, 5).

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