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

ant101 Dec 12, 2018

#1**+1 **

Solve the following system:

{x + 3 y + 2 z = 1 | (equation 1)

-3 x + y + 5 z = 10 | (equation 2)

-2 x - 3 y + z = 7 | (equation 3)

Swap equation 1 with equation 2:

{-(3 x) + y + 5 z = 10 | (equation 1)

x + 3 y + 2 z = 1 | (equation 2)

-(2 x) - 3 y + z = 7 | (equation 3)

Add 1/3 × (equation 1) to equation 2:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x+(10 y)/3 + (11 z)/3 = 13/3 | (equation 2)

-(2 x) - 3 y + z = 7 | (equation 3)

Multiply equation 2 by 3:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x+10 y + 11 z = 13 | (equation 2)

-(2 x) - 3 y + z = 7 | (equation 3)

Subtract 2/3 × (equation 1) from equation 3:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x+10 y + 11 z = 13 | (equation 2)

0 x - (11 y)/3 - (7 z)/3 = 1/3 | (equation 3)

Multiply equation 3 by 3:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x+10 y + 11 z = 13 | (equation 2)

0 x - 11 y - 7 z = 1 | (equation 3)

Swap equation 2 with equation 3:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x - 11 y - 7 z = 1 | (equation 2)

0 x+10 y + 11 z = 13 | (equation 3)

Add 10/11 × (equation 2) to equation 3:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x - 11 y - 7 z = 1 | (equation 2)

0 x+0 y+(51 z)/11 = 153/11 | (equation 3)

Multiply equation 3 by 11/51:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x - 11 y - 7 z = 1 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Add 7 × (equation 3) to equation 2:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x - 11 y+0 z = 22 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 2 by -11:

{-(3 x) + y + 5 z = 10 | (equation 1)

0 x+y+0 z = -2 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract equation 2 from equation 1:

{-(3 x) + 0 y+5 z = 12 | (equation 1)

0 x+y+0 z = -2 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Subtract 5 × (equation 3) from equation 1:

{-(3 x)+0 y+0 z = -3 | (equation 1)

0 x+y+0 z = -2 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

Divide equation 1 by -3:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = -2 | (equation 2)

0 x+0 y+z = 3 | (equation 3)

**x = 1 y = -2 z = 3**

Guest Dec 12, 2018

#3**0 **

Wow, guest. That is a lot of work. I think there is a more "compact" solution. (see below)

PartialMathematician
Dec 13, 2018

#2**0 **

Add eq 1 and 3

-x + 3z = 8 (4)

Multiply eq 2 by 3 and add it to eq 3

-11x +16z=37

Multiply (4) by -11 and add to (5)

-17z= -58 so z=3 Sub into (4) to find x = 1

Sub these into any of the equations to find y

x+3y +2z =1

1 +3y +2(3) = 1 thus y = -2

ElectricPavlov Dec 13, 2018