6 3 Solving Systems Using Elimination Combination Hubarth

6 -3 Solving Systems Using Elimination (Combination) Hubarth Algebra

Ex 1 Adding Equations Solve by elimination. 2 x + 3 y = 11 – 2 x + 9 y = 1 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 Choose the first equation. 2 x + 3(1) = 11 Substitute 1 for y. 2 x + 3 = 11 Solve for x. 2 x = 8 x=4 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 – 8 + 9 1 1 = 1 Substitute 4 for x and 1 for y into the second equation.

Ex 2 Application 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 Relate: total number at the game Write: a + s = 1139 total amount collected 5 a +1 s = 3067 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.

Ex 3 Multiplying One Equation Solve by elimination. 3 x + 6 y = – 6 – 5 x – 2 y = – 14 Step 1: Eliminate one variable. Start with the given system. 3 x + 6 y = – 6 – 5 x – 2 y = – 14 To prepare to eliminate y, multiply the second equation by 3. 3 x + 6 y = – 6 3(– 5 x – 2 y = – 14) Add the equations to eliminate y. 3 x + 6 y = – 6 – 15 x – 6 y = – 42 – 12 x + 0 = – 48 Step 2: Solve for x. – 12 x = – 48 x= 4 Step 3: Solve for the eliminated variable using either of the original equations. 3 x + 6 y = – 6 Choose the first equation. 3(4) + 6 y = – 6 Substitute 4 for x. 12 + 6 y = – 6 Solve for y. 6 y = – 18 y = – 3 The solution is (4, – 3).

Ex 4 Multiplying Both Equations Solve by elimination. 3 x + 5 y = 10 5 x + 7 y = 10 Step 1: Eliminate one variable. Start with the given system. 3 x + 5 y = 10 5 x + 7 y = 10 Step 2: Solve for y. 4 y = 20 y = 5 To prepare to eliminate x, multiply one equation by 5 and the other equation by 3. 5(3 x + 5 y = 10) 3(5 x + 7 y = 10) Subtract the equations to eliminate x. 15 x + 25 y = 50 15 x + 21 y = 30 0 + 4 y = 20 Step 3: Solve for the eliminated variable x using either of the original equations. 3 x + 5 y = 10 Use the first equation. 3 x + 5(5) = 10 Substitute 5 for y. 3 x + 25 = 10 3 x = – 15 x = – 5 The solution is (– 5, 5).

Practice 1. Solve by Elimination a. x + y = 10 x–y =8 b. -2 x + 15 y = -32 7 x – 5 y = 17 c. 15 x + 3 y = 9 10 x + 7 y = -4 (9, 1) (1, -2) 2. Your class sells a total of 64 tickets to a play. A student tickets cost $1, and an adult ticket Costs $2. 50. Your class collects $109 in total ticket sales. How many adult tickets did you sell? How many student tickets did you sell? 30 adults, 34 students