Lesson 3-6: Solving Systems of Linear Equations in Three Variables

Transcription

1 Ok, here we are. The grand finale for linear systems! We are going to add another equation to our linear system and another variable. We will now have three equations with three variables. Oh boy, here we go. What a mess! Take a look at the following linear system. It is a system of linear equations in three variables: x y 3z x 2y z Yikes! Where did the z come from? It is simply a third unknown a third variable. Suffice it to say that many problems in real-world physics or in electrical engineering will come up with a system with three or more variables very similar to the above. How do we solve this? Remember I ve been saying over and over that combination is sometimes the only method you can use? Well this is it. With a linear system in three variables, you can only use combination. All you do is start taking them two at a time, working to eliminate variables: 1. Combine any two of the equations to eliminate one variable. Gives a new two variable equation save it. 2. Combine the third equation with either of the two used in step one to eliminate the same variable. Gives a new two variable equation with the same two variables left in step one save it. 3. Take the two new 2 variable equations from steps 1 and 2. Use combination to solve them for the two variables. 4. Plug the values for those two variables back into one of the original three variable equations to find the remaining variable! Page 1 of 6

2 Let s try it Here is our system again. Look at it and decide which two equations you want to start with so you can eliminate one of the three variables: x y 3z x 2y z Step one: What about the 1 st two equations? If we just add them together, the x s will be eliminated! Let s do it: x y 3z 3y z 1st newequation with 2 var... saveit Step two: Now we need to work with the 3 rd equation and try to eliminate x in it. Why don t we combine the 3 rd and the 1 st? Multiply the 3 rd by 3 and add it with the 1st: x 6 y 3z this is the3rd equation multiplied by 3 4y z 2 nd newequation... Step three: Take the two new 2 variable equations and solve them: 3y z Thetwo... 4y z... newequations 3y z 4 y z 1 2nd equation multiplied by 1 7y y 3(0) z Plug y back in one of these equations z 1 Step four: Plug y = 0 and z = 1 into either of the original equations: x 2y z x 2(0) 1 x 1 0 x 1 There we have it! The solution is (-1, 0, 1). Page 2 of 6

3 That wasn t so bad! Let s try another! Solve the following linear system: x 4y 2z 1 x 2y z 6 2x y 3z 7 Step one: Pick two equations so we can eliminate a variable. Why don t we combine the 1 st two to eliminate z? Multiply the 2 nd by 2 and add to the 1 st : x 4y 2z 1 2x 4 y 2z 12 the 2nd equation multiplied by 2 x 8y 13 we' ve cancelled the z ' s... our1st newequation Step two: We need to use the other original equation and eliminate z again to get a 2 nd two variable equation. Which of the 1 st two equations should we combine with the 3 rd to eliminate z? I ll pick the 2 nd since it s z coefficient is -1 it will be much easier to work with. Just multiple the 2 nd by 3 and add to the 3 rd : 3x 6 y 3z 18 the 2nd equation multiplied by 3 2x y 3z 7 5x 5 y 25 hey look, every termis divisibleby 5... x y 5... divide every termby 5 to simplify! Our 2 nd newequation. Step three: Solve the two new equations that only have x and y as a linear system: x 8y 13 x y 5 9y 18 y 2 We nowknowthevalueof y... x plug y back in and solve for x. x 3 Step four: Plug x = 3 & y = 2 back into one of the original 3 variable equations to find z: x 2y z 6 3 2(2) z 6 7 z 6 z z 1 Our solution is (3, 2, 1). Page 3 of 6

4 Are there any gotchas? Well, yup there are. And here they are Do you remember how many solutions a linear system of parallel lines has? Since the solution of a linear system is the intersection of the lines, if the lines are parallel, there are no solutions. Ok, what about if all the equations in the system turn out to be for the same line? How many solutions do we have then? Since all the lines share all their points, there are an infinite number of solutions. All that applies if there are two equations in the system or there are three. Super, now what does that look like when you solve a three equation linear system? Let s see! I ll show you the two different situations on the next two pages Page 4 of 6

5 A three equation system of parallel lines Solve this linear system: x y 2z 5 2x 3y z 1 Step one: Let s combine the 1 st two to eliminate x: multiply 1st by -1 and add to the 2 nd : 1x y 2z 5 1st equation multiplied by 1 Step two: Let s combine the 1 st & 3 rd to eliminate x: multiply 1 st by -2 and add to the 3 rd : 2x 2 y 4z 0 1st equation multiplied by 2 2x 3y z 1 y 3z 9 Step three: Combine the 2 new equations to eliminate one of the variables. Let s multiply the 2 nd by -1 to eliminate the y s: y z the two... y 3z 9... newequations... y 3z 1st multiplied by 1... y 3z Hmm that s weird. We got 0 = -9 for an answer. Obviously 0 does not equal -9. Anytime you solve a linear system and get one number equal to a different number, this means the lines are all parallel. Hence there are no solutions. Page 5 of 6

6 A three equation system of the same line Solve this linear system: x y 2z 5 2x 3y z 13 Step one: Combine 1 st & 2 nd equations to eliminate x: multiply 1 st by -1 & add to the 2 nd : 1x y 2z 5 1st multiplied by 1 Step two: Combine 1 st & 3 rd equations to eliminate x: multiply 1 st by -2 & add to the 3 rd : 2x 2 y 4z 0 1st equation multiplied by 2 2x 3y z 13 Step three: Combine the 2 new equations to eliminate one of the variables. Let s multiply the 2 nd by -1 to eliminate the y s: y z the two... y z... newequations... y 3z 1st multiplied by Again, this is not normal. We tried to eliminate one variable and ended up getting rid of all variables. This time we ended with a number equals itself; which is true. Anytime you solve a linear system and get a number equal to itself, this means the equations are all for the same line. Hence there are an infinite number of solutions. Page 6 of 6