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5.2 Solving Systems of Linear Equations by Substitution

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1 5.2 Solving Systems of Linear Equations by Substitution

2 Solving Linear Systems by Substitution

3 Example 1: Solving a System of Linear Equations by Substitution
Solve the system of linear equations by substitution. y = -2x – 9 Equation 1 6x – 5y = -19 Equation 2 Step 1: Equation 1 is already solved for y. Step 2: Substitute -2x – 9 in for y in Equation 2 and solve for x. 6x – 5(-2x – 9) = -19 6x + 10x + 45 = -19 16x + 45 = -19 16x = -64 x = -4 Step 3: Substitute -4 in for x in Equation 1 and solve or y. y = -2(-4) – 9 y = 8 – 9 y = -1 The solution is (-4,-1). Remember to check your work by substituting the values back in the original equations!

4 You try! Solve the system of linear equations by substitution. Check your solution! y = 3x ) x = 6y – 7 y = -4x x + y = -3 (-2,8) (-1,1)

5 Example 2: Solving a System of Linear Equations by Substitution
Solve the system of linear equations by substitution. -x + y = 3 Equation 1 3x + y = -1 Equation 2 Step 1: Solve Equation 1 for y. y = x +3 Step 2: Substitute x + 3 in for y in Equation 2 and solve for x. 3x + (x + 3) = -1 4x + 3 = -1 4x = -4 x = -1 Step 3: Substitute -1 in for x in Equation 1 and solve or y. -(-1) + y = 3 1 + y = 3 y = 2 The solution is (-1,2). Remember to check your work by substituting the values back in the original equations!

6 You try! 3) –x + y = -4 4x – y = 10 (2,-2)

7 Example 3: Solving Real-Life Problems
A drama club earns $1040 from a production. A total of 64 adult tickets and 132 student tickets are sold. An adult ticket cost twice as much as a student ticket. Write a system of linear equations to represent the situation. What is the cost of each ticket type? 64∙𝑎𝑑𝑢𝑙𝑡 𝑡𝑖𝑐𝑘𝑒𝑡 𝑝𝑟𝑖𝑐𝑒+132∙𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑡𝑖𝑐𝑘𝑒𝑡 𝑝𝑟𝑖𝑐𝑒=1040 𝐴𝑑𝑢𝑙𝑡 𝑡𝑖𝑐𝑘𝑒𝑡 𝑝𝑟𝑖𝑐𝑒=2∙𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑡𝑖𝑐𝑘𝑒𝑡 𝑝𝑟𝑖𝑐𝑒 Let x be the price (in dollars) of an adult ticket. Let y be the price (in dollars) of a student ticket. 𝑆𝑦𝑠𝑡𝑒𝑚: 𝑥+132𝑦=1040 x = 2y

8 𝑆𝑦𝑠𝑡𝑒𝑚: 64𝑥+132𝑦=1040 Equation 1 x = 2y Equation 2
Step 1: Equation 2 is already solved for x. Step 2: Substitute 2y in for x in Equation 1 and solve for y. 64(2y) + 132y = y + 132y = y = 1040 y = 4 Step 3: Substitute 4 in for y in Equation 2 and solve or x. x = 2(4) x = 8 The solution is (8,4). This means that an adult ticket cost $8 and a student ticket cost $4. Remember to check your work by substituting the values back in the original equations!

9 You try! 4) There are a total of 64 students in a drama club and a yearbook club. The drama club has 10 more students than the yearbook club. Write a system of linear equations that represents this situation. How many students are in each club? # 𝑜𝑓 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑟𝑎𝑚𝑎 𝑐𝑙𝑢𝑏 + # 𝑜𝑓 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟𝑏𝑜𝑜𝑘 𝑐𝑙𝑢𝑏 =64 #𝑜𝑓 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑑𝑟𝑎𝑚𝑎 𝑐𝑙𝑢𝑏 −10=# 𝑜𝑓 𝑠𝑡𝑢𝑑𝑒𝑛𝑡𝑠 𝑖𝑛 𝑡ℎ𝑒 𝑦𝑒𝑎𝑟𝑏𝑜𝑜𝑘 𝑐𝑙𝑢𝑏 Let x be the number of students in the drama club. Let y be the number of students in the yearbook club. The solution is (37,27). This means that there are 37 students in the drama club and 27 students in the yearbook club. 𝑆𝑦𝑠𝑡𝑒𝑚: 𝑥+𝑦=64 x -10 = y


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