Solve Linear Systems by Substitution

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7.2 Solve Linear Systems by Substitution
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Presentation transcript:

Solve Linear Systems by Substitution Warm Up Lesson Presentation Lesson Quiz

Warm-Up Solve the equation. 1. 6a – 3 + 2a = 13 2. 4(n + 2) – n = 11 ANSWER a = 2 ANSWER n = 1 3. You burned 8 calories per minute on a treadmill and 10 calories per minute on an elliptical trainer for a total of 560 calories in 60 minutes. How many minutes did you spend on each machine? ANSWER treadmill: 20 min, elliptical trainer: 40 min

Solve the linear system: y = 3x + 2 Example 1 Solve the linear system: y = 3x + 2 Equation 1 x + 2y = 11 Equation 2 SOLUTION STEP 1 Solve for y. Equation 1 is already solved for y.

Substitute 3x + 2 for y in Equation 2 and solve for x. Example 1 STEP 2 Substitute 3x + 2 for y in Equation 2 and solve for x. x + 2y = 11 Write Equation 2. x + 2(3x + 2) = 11 Substitute 3x + 2 for y. 7x + 4 = 11 Simplify. 7x = 7 Subtract 4 from each side. x = 1 Divide each side by 7.

Example 1 STEP 3 Substitute 1 for x in the original Equation 1 to find the value of y. y = 3x + 2 = 3(1) + 2 = 3 + 2 = 5 ANSWER The solution is (1, 5).

Substitute 1 for x and 5 for y in each of the original equations. Example 1 CHECK Substitute 1 for x and 5 for y in each of the original equations. y = 3x + 2 x + 2y = 11 5 = 3(1) + 2 ? 1 + 2 (5) = 11 ? 5 = 5 11 = 11

Solve the linear system: x – 2y = –6 Example 2 Solve the linear system: x – 2y = –6 Equation 1 4x + 6y = 4 Equation 2 SOLUTION STEP 1 Solve Equation 1 for x. x – 2y = –6 Write original Equation 1. x = 2y – 6 Revised Equation 1

Substitute 2y – 6 for x in Equation 2 and solve for y. Example 2 STEP 2 Substitute 2y – 6 for x in Equation 2 and solve for y. 4x + 6y = 4 Write Equation 2. 4(2y – 6) + 6y = 4 Substitute 2y – 6 for x. 8y – 24 + 6y = 4 Distributive property 14y – 24 = 4 Simplify. 14y = 28 Add 24 to each side. y = 2 Divide each side by 14.

Substitute 2 for y in the revised Equation 1 to find the value of x. Example 2 STEP 3 Substitute 2 for y in the revised Equation 1 to find the value of x. x = 2y – 6 Revised Equation 1 x = 2(2) – 6 Substitute 2 for y. x = –2 Simplify. ANSWER The solution is (–2, 2).

Substitute –2 for x and 2 for y in each of the original equations. Example 2 CHECK Substitute –2 for x and 2 for y in each of the original equations. Equation 1 Equation 2 4x + 6y = 4 x – 2y = –6 –2 – 2(2) = –6 ? 4(–2) + 6 (2) = 4 ? –6 = –6 4 = 4

Guided Practice Solve the linear system using the substitution method. y = 2x + 5 1. ANSWER (1, 7) 3x + y = 10 x – y = 3 2. ANSWER (0, –3) x + 2y = –6 3x + y = –7 3. ANSWER (–2, –1) –2x + 4y = 0

Example 3 WEBSITES Many businesses pay website hosting companies to store and maintain the computer files that make up their websites. Internet service providers also offer website hosting. The costs for website hosting offered by a website hosting company and an Internet service provider are shown in the table. Find the number of months after which the total cost for website hosting will be the same for both companies.

Example 3 SOLUTION STEP 1 Write a system of equations. Let y be the total cost after x months. Equation 1: Internet service provider y = 10 + 21.95 x

Equation 2: Website hosting company Example 3 Equation 2: Website hosting company y = 22.45 x The system of equations is: y = 10 + 21.95x Equation 1 y = 22.45x Equation 2

Substitute 22.45x for y in Equation 1 and solve for x. Example 3 STEP 2 Substitute 22.45x for y in Equation 1 and solve for x. y = 10 + 21.95x Write Equation 1. 22.45x = 10 + 21.95x Substitute 22.45x for y. 0.5x = 10 Subtract 21.95x from each side. x = 20 Divide each side by 0.5. The total cost will be the same for both companies after 20 months. ANSWER

Guided Practice 4. In Example 3, what is the total cost for website hosting for each company after 20 months? $449 ANSWER 5. WHAT IF? In Example 3, suppose the Internet service provider offers $5 off the set-up fee. After how many months will the total cost for website hosting be the same for both companies? 10 mo ANSWER

Example 4 ANTIFREEZE For extremely cold temperatures, an automobile manufacturer recommends that a 70% antifreeze and 30% water mix be used in the cooling system of a car. How many quarts of pure (100%) antifreeze and a 50% antifreeze and 50% water mix should be combined to make 11 quarts of a 70% antifreeze and 30% water mix? SOLUTION STEP 1 Write an equation for the total number of quarts and an equation for the number of quarts of antifreeze. Let x be the number of quarts of 100% antifreeze, and let y be the number of quarts of a 50% antifreeze and 50% water mix.

Example 4 Equation 1: Total number of quarts x + y = 11 Equation 2: Number of quarts of antifreeze x quarts of 100% antifreeze y quarts of 50%–50% mix 11 quarts of 70%–30% mix 1 x + 0.5 y = 0.7(11) x + 0.5y = 7.7

The system of equations is: x + y =11 Example 4 The system of equations is: x + y =11 Equation 1 x + 0.5y = 7.7 Equation 2 STEP 2 Solve Equation 1 for x. x + y = 11 Write Equation 1 x = 11 – y Revised Equation 1 STEP 3 Substitute 11 – y for x in Equation 2 and solve for y. x + 0.5y = 7.7 Write Equation 2.

Substitute 6.6 for y in the revised Equation 1 to find the value of x. Example 4 (11 – y) + 0.5y = 7.7 Substitute 11 – y for x. Solve for y. y = 6.6 STEP 4 Substitute 6.6 for y in the revised Equation 1 to find the value of x. x = 11 – y = 11 – 6.6 = 4.4 ANSWER Mix 4.4 quarts of 100% antifreeze and 6.6 quarts of a 50% antifreeze and 50% water mix to get 11 quarts of a 70% antifreeze and 30% water mix.

Guided Practice WHAT IF? How many quarts of 100% antifreeze and a 50% antifreeze and 50% water mix should be combined to make 16 quarts of a 70% antifreeze and 30% water mix? 6. ANSWER 6.4 quarts of 100% antifreeze and 9.6 quarts of a 50% antifreeze and 50% water mix

Lesson Quiz Solve the linear system using substitution 1. –5x – y = 12 3x – 5y = 4 ANSWER (–2, –2) 2. 2x + 9y = –4 x – 2y = 11 ANSWER (7, –2)

Lesson Quiz 3. You are making 6 quarts of fruit punch for a party. You want the punch to contain 80% fruit juice. You have bottles of 100% fruit juice and 20% fruit juice. How many quarts of 100% fruit juice and how many quarts of 20% fruit juice should you mix to make 6 quarts of 80% fruit juice? ANSWER 4.5 quarts of 100% fruit juice and 1.5 quarts of 20% fruit juice