Intermediate Algebra (11th edition)

(Marvins-Underground-K-12) #1
Solving a System in Three Variables

Solve the system.


(1)
(2)
(3)

Step 1 Since zin equation (1) has coefficient 1, we choose zas the focus variable and


(1) as the working equation. (Another option would be to choose xas the focus


variable, since it also has coefficient 1, and use (2) as the working equation.)


Focus variable
(1) Working equation

Step 2 Multiply working equation (1) by 3 and add the result to equation (2) to


eliminate focus variable z.


Multiply each side of (1) by 3.
(2)
Add. (4)

Step 3 Multiply working equation (1) by and add the result to remaining


equation (3) to again eliminate focus variable z.


Multiply each side of (1) by
(3)
Add. (5)

Step 4 Write the equations in two variables that result in Steps 2 and 3 as a system.


(4) The result from Step 2

- 6 x- 19 y=- 1 (5) The result from Step 3


13 x+ 31 y=- 8


- 6 x- 19 y =- 1


2 x- 3 y+ 2 z = 3


- 8 x- 16 y- 2 z =- 4 - 2.


- 2


13 x+ 31 y =- 8


x+ 7 y- 3 z=- 14


12 x+ 24 y+ 3 z= 6


4 x+ 8 y+z= 2


2 x- 3 y+ 2 z= 3


x+ 7 y- 3 z=- 14


4 x+ 8 y+ z= 2


EXAMPLE 1


228 CHAPTER 4 Systems of Linear Equations


Make sure these
equations have the
same variables.

Now solve this system. We choose to eliminate x.


Multiply each side of (4) by 6.
Multiply each side of (5) by 13.
Add.
Divide by

Substitute 1 for yin either equation (4) or (5) to find x.


(5)
Let
Multiply.
Add 19.
Divide by

Step 5 Now substitute the two values we found in Step 4 in working equation (1) to


find the value of the remaining variable, focus variable z.


(1)
Let
Multiply, and then add.

z= 6 Add 4.


- 4 +z= 2


41 - 32 + 8112 +z= 2 x=-3 and y=1.


4 x+ 8 y+z= 2


x= - 3 - 6.


- 6 x= 18


- 6 x- 19 =- 1


- 6 x- 19112 =- 1 y=1.


- 6 x- 19 y=- 1


y= 1 - 61.


- 61 y=- 61


- 78 x- 247 y=- 13


78 x+ 186 y=- 48

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