Yingda Z. answered • 09/16/13

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Hi Courtnee,

It is a good question to cast doubt if there is a unique solution to your linear system (we call it linear because all three equation you listed here are in linear form , i.e ax+by+cz=k where a,b,c,k are numbers and x,y,z are unknows).

The basic idea to solve a linear system is to use the method called "elimination of variables" , i.e. multiplying a specific number to one of your equation and add to another to reduce the numbers of unknowns. For your case, it looks like this:

1). A good start point is observing that the first and third equation look similar. (and they do!). Try to add first equation to the third you get

(x-x)-(2y+2y)+(z-z)=-6+6

which ends up with a equivalence "0=0" which means that adding first and third equation give you no information about the value of x,y and z because for any x,y and z, "(x-x)-(2y+2y)+(z-z)=-6+6" always hold!

In another word, the first and third equation gives you the same information and you can actually drop one of them without harming the solution power of the system. Let's drop the third one.

2). Thus you only have two equations: x-2y+z=-6; and 3x-6y+2z=12. As you notice, you have three unknowns but two equations. You can conclude now that there's no unique solution to this linear system. Even this is the case, you can still go a step further. Multiplying (-2) on the first equation and then add it to the second one, we get

(-2x+3x) + (4y-6y) + (-2z+2z)=12+12 (*)

The reason we multiplying (-2) on the first one and add it to the second is to eliminate the unknown "z".

Then equation (*) ends up with

x-2y=24 (**)

For a prealgebra (or maybe algebra 1) course, you can stop here and say "there is no unique solution to the system and any x,y satisfying (**) and x,y,z satisfying the first equation (or third one) is a pair of solution."

You can ignore what I am going to say if you are not interested or already feel overwhelmed with the information. Otherwise you may learn that you can actually say more than that by treating one of the unknown as "independent variable". You can pick any one of the three as the independent variable in the linear system, let's say "y". As the name suggests, we can use "y" to express the other two variable x and z and we dont know the specific value of y. How to do this ?

3). From (**), rearrange and get x=24+2y; From the first or third equation you get z=-6+2y-x=-6+2y-(24+2y)=-6+2y-2y-24=-30. You can actually solve for z=-30! But unfortunately, you still cant solve for specific value for x and y. Finally you can say that the solution to this linear system is the following

x=2y+24 and z=-30 given any value of y.

Generally speaking, solving linear system is the #1 issue in linear algebra when you go to college math class. The reason why there's no unique solution is because the first and the third equations are not "linearly independent" giving you the same information. Hence the system looks like you have three equations and three unknowns but it actually does not. After observing the fact, what you have is only two equations with three unknowns.

Hope it helps you to solve the other question you posted!

Best

Yingda Z.