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Find the ordered triplet \((x,y,z)\)for the following system of equations:

\(\begin{align*} x + 3y + 2z &= 1,\\ -3x + y + 5z &= 10,\\ -2x - 3y +z &= 7. \end{align*}\)

Include the parentheses for the ordered triplet in your answer.

 Oct 27, 2019
 #1
avatar+106539 
+1

x + 3y + 2z  =  1

-3x + y + 5z  =  10

-2x - 3y + z  =  7

 

Add  the first and third equations ⇒   -x + 3z  =  8   

Multiply this by 10   ⇒  -10x + 30z  =  80     (4) 

 

Multiply the second equation by  -3  ⇒   9x -3y  - 15z  = -30

Add this to the first equation ⇒    10x  -13z = -29    (5)

 

Add (4)  and (5)   ⇒  17z  =  51

Divide both sides by  17

z  =  3

 

And 

-x + 3(3)  =  8

-x  +  9  = 8

-x  = -1

x  = 1

 

And

1 + 3y + 2(3)  = 1

1 + 3y + 6  = 1

3y + 6  = 0

3y  = 6

y = - 2

 

So

 

{x, y, z}   =  { 1, - 2, 3 }

  

 

 

cool cool cool

 Oct 27, 2019
 #2
avatar+308 
+3

First, we add the first and third equations to get -x+3z=8. Then, we multiply the first equation by two and add the third equation to that to get 3y+5z=9. Then, we subtract that from the second equation to get -3x-2y=1. Then, we add the first equation to the second and subtract the third equation from that to get 7y+6z=4. This means that we now have the equations

 

-x+3z=8

3y+5z=9

-3x-2y=1

7y+6z=4

 

Looking at the second and fourth equations, we find that y = -2 and that z = 3. Solving for x, we find that x = 1. I'll leave the checking up to you.

 Oct 27, 2019
 #3
avatar+106539 
+1

THX, ThatOnePerson   !!!!

 

 

cool cool cool

CPhill  Oct 27, 2019

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