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.

Guest Oct 27, 2019

#1**+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 }

CPhill Oct 27, 2019

#2**+2 **

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.

ThatOnePerson Oct 27, 2019