Lesson 6-3 Warm-Up

“Solving Systems Using Elimination” (6-3) What is the elimination method for solving a system of equations? Method 3: Elimination Method: If the coefficients for one of the terms in both equations has a sum or difference of zero (in other words, the terms are opposites, like 6x and -6x or d and -d), you can use The Addition or Subtraction Properties of Equality to eliminate the variable. Then, you can solve for the variable that is not eliminated and use substitution to find the other variable. Example: Solve 5x – 6y = -32 and 3x + 6y = 48 Step 1: Add the equations to eliminate the 6y and -6y (opposite terms) and solve for x. Step 2: Solve for y by substituting the solution for x into one of the original equations. Since x = 2 and y = 7, the solution is (2, 7) Step 3: Check if the solution makes both equations true statements 5x – 6y = -32 3x + 6y = 48 Start with original equations. 5(2) - 6(7) = – 33 3(2) + 6(7) = 48 y = 7 and x = 2 (substitute) 32 = 32 48 = 48

Solve by elimination. 2x + 3y = 11 –2x + 9y = 1 Solving Systems Using Elimination LESSON 6-3 Additional Examples Solve by elimination. 2x + 3y = 11 –2x + 9y = 1 Step 1: Eliminate x because the sum of the coefficients is 0. 2x + 3y = 11 –2x + 9y =1 12y = 12 Addition Property of Equality y = 1 Solve for y. 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

Since x = 4 and y = 1, the solution is (4, 1). Solving Systems Using Elimination LESSON 6-3 Additional Examples (continued) Since x = 4 and y = 1, the solution is (4, 1). Check: See if (4, 1) makes the equation not used in Step 2 true. –2(4) + 9(1) 1 Substitute 4 for x and 1 for y into the second equation. –8 + 9 1 1 = 1

Define: Let a = number of adults Let s = number of students Solving Systems Using Elimination LESSON 6-3 Additional Examples On a special day, tickets for a minor league baseball game cost $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. Define: Let a = number of adults Let s = number of students Words: total number at the game total amount collected Equation: a + s = 1139 5 a + s = 3067 Since you have an “s” in both equations, solve by subtracting the equations

Step 1: Eliminate one variable. a + s = 1139 (-5a) + (-s) = (-3067) Solving Systems Using Elimination LESSON 6-3 Additional Examples (continued) Step 1: Eliminate one variable. a + s = 1139 (-5a) + (-s) = (-3067) –4a = –1928 Subtraction Property of Equality a = 482 Solve for a. -1(5a + s = 3067) = [-5a + (-s) = -3067] + 0 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. -482 -482 Subtract 482 from both sides. 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.

“Solving Systems Using Elimination” (6-3) How can you eliminate one of the variables if the sum or difference of one of the variables doesn’t equal 0? Sometimes, you may need to multiply one or both equations in a system of equations by different nonzero numbers in order to eliminate one of the variables by adding or subtracting the equations. Example: Solve 4x + 2y = 14 and 7x - 3y = -8 Step 1: We see that the y coefficients, 2 and 3 can become opposites if we multiplied them by each other. Therefore, we’ll multiply the equation with 2y by 3 and the equation with -3y by 2 to create equivalent equations with opposite terms. Then, we can add the two equivalent equations together to eliminate one of the equations. Step 2: Solve for x. 26x = 26 Step 1 26 26 Divide both sides by 26. x = 1

“Solving Systems Using Elimination” (6-3) Step 3: Solve for the eliminated variable, y, by substituting the solution for x into one of the original equations. 4x + 2y = 14 Given 4(1) + 2y = 14 Substitute (x = 1) -4 -4 Subtract 4 from both side 2y = 10 Simplify 2 2 Divide both side by 2 y = 5 Since x = 1 and y = 5, the solution is (1, 5) Step 4: Check if the solution makes both equations true statements 4(1) + 2(5) = 14 7(1) – 3(5) = -8 4 + 10 = 14 7 - 15 = -8 y = 7 and x = 2 (substitute) 14 = 14 -8 = -8

Solve by elimination. 3x + 6y = –6 –5x – 2y = –14 Solving Systems Using Elimination LESSON 6-3 Additional Examples 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 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 Step 2: Solve for x. –12x = –48 x = 4 Divide both sides by -12.

3x + 6y = –6 Choose the first equation. Solving Systems Using Elimination LESSON 6-3 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 Subtract 12 from both sides. 6y = –18 Divide both sides by 6. y = –3 The solution is (4, –3).

Define: Let p = number of cans of popcorn sold. Solving Systems Using Elimination LESSON 6-3 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. Define: Let p = number of cans of popcorn sold. Let n = number of cans of nuts sold. Words: total number of cans total amount of sales Equations: p + n = 240 5 p + 8 n = 1614

Step 1: Eliminate one variable. Solving Systems Using Elimination LESSON 6-3 Additional Examples (continued) Step 1: Eliminate one variable. 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 Step 2: Solve for n. –3n = –414 Divide both sides by -3. n = 138

Step 3: Solve for the eliminated variable using either of the original Solving Systems Using Elimination LESSON 6-3 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.

Solve by elimination. 3x + 5y = 10 5x + 7y = 10 Solving Systems Using Elimination LESSON 6-3 Additional Examples Solve by elimination. 3x + 5y = 10 5x + 7y = 10 Step 1: Eliminate one variable. 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 Step 2: Solve for y. 4y = 20 Divide both sides by 4. y = 5

3x + 5y = 10 Use the first equation. Solving Systems Using Elimination LESSON 6-3 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 Subtract 25 from both sides. 3x = –15 x = –5 The solution is (–5, 5).

Solve using elimination. 1. 3x – 4y = 7 2. 5m + 3n = 22 Solving Systems Using Elimination LESSON 6-3 Lesson Quiz Solve using elimination. 1. 3x – 4y = 7 2. 5m + 3n = 22 2x + 4y = 8 5m + 6n = 34 3. –6x + 5y = 4 4. 7p + 5q = 2 3x + 4y = 11 8p – 9q = 17 (3, 0.5) (2, 4) (1, 2) (1, 1)