Solving Systems Using Elimination Solve each system using

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

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

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

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

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

Solving Systems Using Elimination ALGEBRA 1 LESSON 9 -4 (continued) Since x = 4 and y = 1, the solution is (4, 1). 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 9 -4

Solving Systems Using Elimination ALGEBRA 1 LESSON 9 -4 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. Define: Let a = number of adults Let s = number of students Relate: total number at the game Write: a + s = 1139 total amount collected 5 a Solve by elimination. 9 -4 + s = 3067

Solving Systems Using Elimination ALGEBRA 1 LESSON 9 -4 (continued) Step 1: Eliminate one variable. a + s = 1139 5 a + s = 3067 – 4 a + 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. 9 -4

Solving Systems Using Elimination ALGEBRA 1 LESSON 9 -4 Solve using elimination. 1. 3 x – 4 y = 7 2 x + 4 y = 8 (3, 0. 5) 2. 5 m + 3 n = 22 (2, 4) 5 m + 6 n = 34 9 -4