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Advanced Mathematical Concepts Chapter 2 Lesson 2-2 Example 1 Solve the system of equations by elimination. x + 4y – z = 20 3x + 2y + z = 8 2x – 3y + 2z = -16 One way to solve a system of three equations is to choose pairs of equations and then eliminate one of the variables. Because the coefficient of x is 1 in the first equation, it is a good choice for eliminating x from the second and third equations. To eliminate x using the first and second To eliminate x using the first and third equations, multiply each side of the first equations, multiply each side of the first equation by –3. equation by –2. -3(x + 4y – z) = -3(20) -3x – 12y + 3z = -60 -2(x + 4y – z) = -2(20) -2x – 8y + 2z = -40 Then add that result to the second equation. -3x – 12y + 3z = -60 3x + 2y + z = 8 -10y + 4z = -52 Then add that result to the third equation. -2x – 8y + 2z = -40 2x – 3y + 2z = -16 -11y + 4z = -56 Now you have two linear equations in two variables. Solve this system. Eliminate z by multiplying each side of the second equation by –1 and adding the two equations. -10y + 4z = -52 -1(-11y + 4z) = -1(-56) -10y + 4z = -52 11y – 4z = 56 y = 4 The value of y is 4. By substituting the value of y into one of the equations in two variables, we can solve for the value of z. -10y + 4z = -52 -10(4) + 4z = -52 y=4 4z = -12 z = -3 The value of z is –3. Finally, use one of the original equations to find the value of x. x + 4y – z = 20 x + 4(4) – (-3) = 20 y = 4, z = -3 x =1 The solution is x = 1, y = 4, and z = -3. This can be written as the ordered triple (1, 4, -3). Check by substituting the values into each of the original equations. Example 2 Solve the system of equations by substitution. 2x = -6y x + y + z = 10 -4x – 4y – z = -4 You can easily solve the first equation for x. 2x = -6y x = -3y Divide each side by 2. Then substitute –3y for x in each of the other two equations. Simplify each equation. x + y + z = 10 -4x – 4y – z = -4 -3y + y + z = 10 x = -3y -4(-3y) – 4y – z = -4 x = -3y -2y + z = 10 8y – z = -4 Solve –2y + z = 10 for z. -2y + z = 10 z = 10 + 2y Add 2y to each side. Substitute 10 + 2y for z in 8y – z = -4. Simplify. 8y – z = -4 8y – (10 + 2y) = -4 z = 10 + 2y Advanced Mathematical Concepts Chapter 2 6y – 10 = -4 y =1 Now, find the values of z and x. Use z = 10 + 2y and x = -3y. Replace y with 1. z = 10 + 2y z = 10 + 2(1) z = 12 x = -3y x = -3(1) x = -3 y=1 y=1 The solution is x = -3, y = 1, and z = 12. Check each value in the original system. Example 3 MANUFACTURING A manufacturer of golf balls supplies three driving ranges with balls. The output from a week’s production of balls is 320 cases. The manufacturer must send driving range A three times as many cases as are sent to driving range B, and must send driving range C 160 cases less than ranges A and B together. How many cases should be sent to each driving range to distribute the entire week’s production to them? Write a system of equations. Define the variables as follows. x = the number of cases sent to driving range A y = the number of cases sent to driving range B z = the number of cases sent to driving range C The system is: x + y + z = 320 x = 3y z = x + y – 160 total number of cases produced cases to A equal three times cases to B cases to C equal 160 less than A and B together The second equation tells us that 3y can be substituted for x in the other two equations. Simplify each equation. x + y + z = 320 3y + y + z = 320 4y + z = 320 z = x + y – 160 z = 3y + y – 160 z = 4y - 160 x = 3y x = 3y Substitute 4y – 160 for z in 4y + z = 320. 4y + z 4y + 4y – 160 8y – 160 y = 320 = 320 = 320 = 60 z = 4y - 160 Now, find the values of x and z. Use x = 3y and z = 4y – 160. Replace y with 60. x = 3y x = 3(60) x = 180 y = 60 z = 4y – 160 z = 4(60) – 160 z = 80 y = 60 The manufacturer should send 180 cases of golf balls to driving range A, 60 cases to driving range B, and 80 cases to driving range C. Assignment: Pages 76-77 8, 11, 14, 17, 24, 25, 28