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Warm-Up Exercises Evaluate for and1. 4.4. y = 3x3x–5y5y – 3x = – Solve the system by graphing.2. 2x+y = 32x2x+y = ANSWER 11 – ANSWER () 1, 1.

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Presentation on theme: "Warm-Up Exercises Evaluate for and1. 4.4. y = 3x3x–5y5y – 3x = – Solve the system by graphing.2. 2x+y = 32x2x+y = ANSWER 11 – ANSWER () 1, 1."— Presentation transcript:

1 Warm-Up Exercises Evaluate for and1. 4.4. y = 3x3x–5y5y – 3x = – Solve the system by graphing.2. 2x+y = 32x2x+y = ANSWER 11 – ANSWER () 1, 1

2 Warm-Up Exercises Twice a number x plus a number y is 3. The number y subtracted from three times the number x is 7. Find x and y by graphing. 3. ANSWER () 2, 1 –

3 Use Substitution Example 1 Solve the system using substitution. 6 = y – 4x4x 2x2x = y SOLUTION Substitute 2x for y in Equation 2. Solve for x. Write Equation 2. 6 = y – 4x4x Substitute 2x for y. 6 = 2x2x – 4x4x 6 = 2x2x Combine like terms. 3 = x Solve for x. Equation 1 Equation 2

4 Use Substitution Example 1 Substitute 3 for x in Equation 1. Solve for y. Write Equation 1. 2x2x = y Substitute 3 for x. () 3 2 = y Solve for y. 6 = y You can check your answer by substituting 3 for x and 6 for y in both equations. ANSWER The solution is. () 3, 6

5 Use Substitution Example 2 Solve the system using substitution. Equation 1 7 = 2y2y3x3x + Equation 2 3 = 2y2y – x – SOLUTION STEP 1 Solve Equation 2 for x. Choose Equation 2 because the coefficient of x is 1. 3 = 2y2y – x – Solve for x to get revised Equation 2. 3 = x – 2y2y

6 Use Substitution Example 2 STEP 2 Substitute 2y 3 for x in Equation 1. Solve for y. – Write Equation 1. 7 = 2y2y3x3x + 7 = 2y2y + 96y6y – Use the distributive property. 7 = 98y8y – Combine like terms. 16 = 8y8y Add 9 to each side. 2 = y Solve for y. 7 = 2y2y3 + () 32y2y – Substitute 2y 3 for x. –

7 Use Substitution Example 2 STEP 3 Substitute 2 for y in revised Equation 2. Solve for x. Write revised Equation 2. 3 = 2y2yx – Substitute 2 for y. 3 = 2x – () 2 Simplify. = 1x STEP 4 Check by substituting 1 for x and 2 for y in the original equations. Equation 1Equation 2 7 = 2y2y3x3x + = 2y2y – x3 – Write original equations.

8 Use Substitution Example 2 73 () 1 +2 () 2 = ? 12 () 2 = ? – 3 – Substitute for x and y. 73+4 = ? 14 = ? – 3 – Simplify. 77 = = 3 – 3 – Solution checks. ANSWER The solution is. () 1, 2

9 Solve the system using substitution. Tell which equation you chose to solve and use for the substitution. Explain. Checkpoint 1. Use Substitution 3 = y2x2x + 0 = y3x3x + ANSWER Sample answer: The second equation; this equation had 0 on one side and the coefficient of y was 1, so I solved for y to obtain y 3x. = – (), 9 3 –.

10 Solve the system using substitution. Tell which equation you chose to solve and use for the substitution. Explain. Checkpoint 2. Use Substitution 4 = 3y3y2x2x + 1 = 2y2yx + ANSWER Sample answer: The second equation; the coefficient of x in this equation was 1, so solving for x gave a result that did not involve any fractions. () 5,5, 2 –.

11 Solve the system using substitution. Tell which equation you chose to solve and use for the substitution. Explain. Checkpoint 3. Use Substitution 10 = 2y2y4x4x + 5 = y3x3x – ANSWER Sample answer: The first equation; the coefficient of y in this equation was 1, so solving for y gave a result that did not involve any fractions. – () 2, 1.

12 Write and Use a Linear System Example 3 Museum Admissions On one day, the Henry Ford Museum in Dearborn, Michigan, admitted 4400 adults and students and collected $57,200 in ticket sales. The price of admission is $14 for an adult and $10 for a student. How many adults and how many students were admitted to the museum that day? SOLUTION VERBAL MODEL Total number admitted Number of adults Number of students =+ = + Total amount collected Student price Adult price Number of adults Number of students

13 Write and Use a Linear System Example 3 LABELS Number of adults x = (adults) Number of students y = (students) Total number admitted 4400 = (dollars) Price for one adult 14 = Price for one student 10 = (people) Total amount collected 57,200 = (dollars) ALGEBRAIC MODEL Equation 1 (number admitted) 4400 = x + y Equation 2 (amount collected) 57,200 = 14x + 10y

14 Write and Use a Linear System Example 3 Use substitution to solve the linear system. 4400 = xy – Solve Equation 1 for x ; revised Equation 1. 57,200 = + 10y14y61,600 – Use the distributive property. 57,200 = 4y4y61,600 – Combine like terms. 57,200 = 14 + 10y () y4400 – Substitute 4400 y for x in Equation 2. – 4400 = 4y4y –– Subtract 61,600 from each side. 1100 = y Divide each side by 4. –

15 Write and Use a Linear System Example 3 ANSWER There were 3300 adults and 1100 students admitted to the Henry Ford Museum that day. 4400 = xy – Write revised Equation 1. 4400 = x1100 – Substitute 1100 for y. 3300 = x Simplify.

16 Checkpoint Write and Use a Linear System 4. On another day, the Henry Ford Museum admitted 1300 more adults than students and collected $56,000. How many adults and how many students were admitted to the museum that day? ANSWER 2875 adults and 1575 students

17 Checkpoint Write and Use a Linear System ANSWER The answers are the same; you can solve for either variable to solve a system. 5. Solve the system of equations in Example 3 by solving Equation 1 for y instead of x. Compare your solution to the solution in Example 3. What conclusion can you make?


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