3.4+Solve+Systems+of+Linear+Equations+in+Three+Variables

**//__ 3.4 Solve Systems of Linear Equation in 3 variables __//**
A __//** linear equation in three variables **//__** is an equation in the form of //__ ax+by+cz=d __// in which //__ a, b, __// and //__ c __//** **are all not zero.** The __**// solution //**__ for such a problem **__is in the form of an ordered triple whose coordinates make each equation true__**. The solution can be found by three steps which are:

1. Rewrite the linear system in three variables as a linear system in two variables by using the elimination method.

2. Solve the linear system for both of it variables.

3. Substitute the values in step two into one of the original equations and solve for the remaining variable.

For **example of how to solve** one of these equations: Click Here

Or, for a math king **video** example: Click Here <advised

For **extra help**, or **more information** on the topic: Click Here = walk through = (sorry for the sketchyness, it isnt copy pasted like yours.) using the elimination method for linear triples
 * If the three planes end up lying on top of each other, then there is an infinite number of solutions. (First image)**
 * If the system in three variables has one solution, it is an ordered triple (//x//, //y//, //z//) that is a solution to ALL THREE equations. (Second image)**
 * If the three planes are parallel to each other, they will never intersect. (Third image)**

4x+2y+3z=1 equation A 2x-3y+5z=-14 equation B 6x-y+4z=-1 equation C


 * step 1. //rewrite//** the system as a linear stystem in two variables.

4x+2y+3z=1 Multiply 2 times equation C ---> 2(6x-y+4z=-1) THEN add to equation A. __12x-2y+8z=-2__
 * 16x+11z=-1 New Equation A.**

2x-3y+5z=-14 multiply -3 times equation C ---> -3(6x-y+4z=-1) THEN add to equation B. __-18x+3y-12z=3__
 * -16x-7z=-11 new equation B**.

16x+11z=-1 add new equation A and new equation B. __-16x-7z=-11__
 * //step 2. Solve the new linear system for both of its variables.//**
 * 4z=-12**
 * z=-3 solve for Z**.
 * Subtitute Z for -3 into equation A or B above and solve for X.**
 * The answer should come out to X=2.**

6x-y+4z=-1 write original equation C. 6(2)-y+4(-3)=-1 substitute 2 for x and -3 for z. y=1 solve for y.
 * //step 3. substitute//** x=2 and z=-3 into an original eqaution and solve for y.
 * the solution is x=2, y=1, and z=-3, or ordered triple (2,1,-3)**

check your solution in each of the original equations.

solve the system x+y+z=3 equation A 4x+4y+4z=7 equation B 3x-y+2z=5 equation C When you muliply Equation A by -4 and add the result to equation B, you obtain a false equation.
 * Solve a three - variable system with no solution**
 * Solution**

-4x-4y-4z=-12 multiply -4 to equation A -4(x+y+z=3) THEN add to equation B __4x+4y+4z=7__
 * 0=5** New equation A

Because you obtain a false equation, you can conclude that the original system has no solution.


 * Solve a three variable system with many solutions**

Solve the system x+y+z=4 equation A x+y-z=4 equation B 3x+3y+z=12 equation C Step 1 Rewrite the system as a linear system in //two// variables.
 * Solution**

x+y+z=4 Add equation A to equation B __x+y-z-=4__ 2x+2y=8 New equation A

x+y-z=4 Add equation B to equation C. __3x+3y+z=12__ 4x+4y=16 New equation B

Step 2 **Solve** the new linear system for both of its variables. -4x-4y=-16 multiply -2 times new equation A -2(2x+2y=8) THEN add to new equation B __4x+4y=16__
 * 0=0**
 * with many solutions x=x will always be the same**
 * Because** you obtain the identity 0=0, the system has infinitely many solutions.

Step 3 Describe the solutions of the system. One way to do this is to divide the new Equation 1 by 2 to get x+y=4 oy y=-x+4. Substituting this into the original Equation 1 produces z=0. So, any ordered triple of the form (x, -x+4, 0) is the solution to the system.