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

Solving Systems Using Elimination 1. y = 4x – 3 y = 2x + 13 Substitute 4x – 3 for y in the second equation. y = 2x x – 3 = 2x x – 2x – 3 = 2x – 2x x – 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 9-4 Solutions 9-4

Solving Systems Using Elimination 2. y + 5x = 4 y = 7x – 20 Substitute 7x – 20 for y in the first equation. y + 5x = 4 7x – x = 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 9-4 Solutions (continued) 9-4

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 9-4 Solutions (continued) 9-4

Solving Systems Using Elimination 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 = y = 12Addition Property of Equality y = 1Solve for y. Step 2:Solve for the eliminated variable x using either original equation. 2x + 3y = 11Choose the first equation. 2x + 3(1) = 11Substitute 1 for y. 2x + 3 = 11 Solve for x. 2x = 8 x = 4 ALGEBRA 1 LESSON

Solving Systems Using Elimination (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) 1Substitute 4 for x and 1 for y into the second equation. – = 1 ALGEBRA 1 LESSON

Define:Let a = number of adults Let s = number of students Relate:total number at the gametotal amount collected Write: a + s = a + s = 3067 Solving Systems Using Elimination 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. ALGEBRA 1 LESSON

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

Solving Systems Using Elimination Solve using elimination. 1.3x – 4y = 72. 5m + 3n = 22 2x + 4y = 8 5m + 6n = 34 (3, 0.5)(2, 4) ALGEBRA 1 LESSON