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Solving Systems of Equations in Three Variables

Solving Systems of Equations in Three Variables. Section 2-2. Before finishing this section you should be able to:. Solve systems of equations involving three variables algebraically.

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Solving Systems of Equations in Three Variables

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  1. Solving Systems of Equations in Three Variables Section 2-2

  2. Before finishing this section you should be able to: • Solve systems of equations involving three variables algebraically Remember: Your textbook is your friend! This presentation is just a supplement to the text. BEFORE you view this, make sure you read this section in your textbook and look at all the great examples that are also worked there for you.

  3. We can solve systems of equations with three variables algebraically with the same methods we used to solve for two variables. The solution is an ordered triple (x, y, z). 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.

  4. To eliminate x using the first and second equations, multiply each side of the first equation by –3. -3(x + 4y – z) = -3(20) -3x – 12y + 3z = -60 Then add that result to the second equation. -3x – 12y + 3z = -60 3x + 2y + z = 8 -10y + 4z = -52

  5. To eliminate x using the first and third equations, multiply each side of the first equation by –2. -2(x + 4y – z) = -2(20) -2x – 8y + 2z = -40 Then add that result to the third equation. -2x – 8y + 2z = -40 2x – 3y + 2z = -16 -11y + 4z = -56

  6. 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.

  7. 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.

  8. 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.

  9. 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…

  10. x + y + z = 10 -3y + y + z = 10 x = -3y -2y + z = 10 -4x – 4y – z = -4 -4(-3y) – 4y – z = -4 x = -3y 8y – z = -4 Solve –2y + z = 10 for z. -2y + z = 10 z = 10 + 2yAdd 2y to each side.

  11. Substitute 10 + 2y for z in 8y – z = -4. Simplify. 8y – (10 + 2y) = -4 z = 10 + 2y 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 + 2yx = -3y z = 10 + 2(1) y = 1x = -3(1) z = 12 x = -3 The solution is x = -3, y = 1, and z = 12 (-3, 1, 12) Check each value in the original system.

  12. Real-World Example 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

  13. The system is: x + y + z = 320 total number of cases produced x = 3ycases to A equal three times cases to B z = x + y – 160 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. z = x + y – 160 z = 3y + y – 160x = 3y z = 4y - 160 x + y + z = 320 3y + y + z = 320 x = 3y 4y + z = 320

  14. Substitute 4y – 160 for z in 4y + z = 320. 4y + z = 320 4y + 4y – 160 = 320 z = 4y - 160 8y – 160 = 320 y = 60 Now, find the values of x and z. Use x = 3y and z = 4y – 160. Replace y with 60. x = 3yz = 4y – 160 x = 3(60) y = 60z = 4(60) – 160 y = 60 x = 180 z = 80 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.

  15. Another Example 5x-2y+z = 11 2x+y+3z = 0 6x-2y-2z = 16 5x-2y+z = 11 2(2x+y+3z) = 2(0) 5x-2y+z = 11 4x+2y+6z = 0 9x +7z = 11 First we must eliminate one of the variables from two of the equations. Let’s get rid of the y-variable. (If any variable in any of the three equations has a coefficient of 1that is going to be the easiest variable to eliminate) To eliminate y using the first and second equations, multiply each side of the second equation by 2. Then add that result to the second equation. Now we have a linear equation in two variables.

  16. Now eliminate y using the second and third equations. Multiply each side of the second equation by 2. Add the result to the third equation. Now we have another linear equation in two variables. 2(2x+y+3z) = 2(0) 6x-2y-2z = 16 4x+2y+6z = 0 6x-2y-2z = 16 10x +4z = 16

  17. Now we have two linear equations in two variables. Solve this system using any of the methods from the last section. (Elimination is used here) Eliminate x by multiplying the each side of the first equation by 10 and each side of the second equation by –9. Add the two equations. The value of z is –1. 9x+7z = 11 10x+4z = 16 10(9x+7z) = 10(11) -9(10x+4z) = -9(16) 90x+70z = 110 -90x-36z = -144 34z =-34 z =-1

  18. By substituting the value of z into one of the equations in two variables, we can solve for the value of x. The value of x=2. Finally, use one of the original equations to find the value of y. The value of y is –1. The answer can be written as an ordered triple (2,-1,-1) 9x+7(-1)=11 9x-7=11 x=2 5(2)-2y+(-1)=11 10-2y-1=11 y=-1

  19. Solving Systems with 3 variables: http://tutorial.math.lamar.edu/AllBrowsers/AlgebraTrigReview/SolveSystems.asp http://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut50_systhree.htm http://fp.academic.venturacollege.edu/rbrunner/3e_3s.htm 2-2 Self-Check Quiz: http://www.glencoe.com/sec/math/studytools/cgi-bin/msgQuiz.php4?isbn=0-07-860861-9&chapter=2&lesson=2&quizType=1&headerFile=4&state= Helpful Websites

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