Solve the following system of linear equations:

x+ 2y+ 3z+ 4u=20

-x + 5 y +2 u=13

x+ y+ z+ u =10

-x+y-z+u= -2

Guest Oct 26, 2015

#1**+10 **

Solve the following system of linear equations:

x+ 2y+ 3z+ 4u=20

-x + 5 y +2 u=13

x+ y+ z+ u =10

-x+y-z+u= -2

Solve the following system:

{4 u+x+2 y+3 z = 20 | (equation 1)

2 u-x+5 y = 13 | (equation 2)

u+x+y+z = 10 | (equation 3)

u-x+y-z = -2 | (equation 4)

Add equation 1 to equation 2:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

x+y+z+u = 10 | (equation 3)

-x+y-z+u = -2 | (equation 4)

Subtract equation 1 from equation 3:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x-y-2 z-3 u = -10 | (equation 3)

-x+y-z+u = -2 | (equation 4)

Multiply equation 3 by -1:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+y+2 z+3 u = 10 | (equation 3)

-x+y-z+u = -2 | (equation 4)

Add equation 1 to equation 4:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+y+2 z+3 u = 10 | (equation 3)

0 x+3 y+2 z+5 u = 18 | (equation 4)

Subtract 1/7 × (equation 2) from equation 3:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+(11 z)/7+(15 u)/7 = 37/7 | (equation 3)

0 x+3 y+2 z+5 u = 18 | (equation 4)

Multiply equation 3 by 7:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+3 y+2 z+5 u = 18 | (equation 4)

Subtract 3/7 × (equation 2) from equation 4:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+(5 z)/7+(17 u)/7 = 27/7 | (equation 4)

Multiply equation 4 by 7:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+5 z+17 u = 27 | (equation 4)

Subtract 5/11 × (equation 3) from equation 4:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+0 z+(112 u)/11 = 112/11 | (equation 4)

Multiply equation 4 by 11/112:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 15 × (equation 4) from equation 3:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+0 u = 22 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Divide equation 3 by 11:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 3 × (equation 3) from equation 2:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+0 z+6 u = 27 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 6 × (equation 4) from equation 2:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+0 z+0 u = 21 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Divide equation 2 by 7:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 2 × (equation 2) from equation 1:

{x+0 y+3 z+4 u = 14 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 3 × (equation 3) from equation 1:

{x+0 y+0 z+4 u = 8 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 4 × (equation 4) from equation 1:

{x+0 y+0 z+0 u = 4 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Collect results:

Answer: |

| {x = 4

y = 3

z = 2

u = 1

Guest Oct 26, 2015

#1**+10 **

Best Answer

Solve the following system of linear equations:

x+ 2y+ 3z+ 4u=20

-x + 5 y +2 u=13

x+ y+ z+ u =10

-x+y-z+u= -2

Solve the following system:

{4 u+x+2 y+3 z = 20 | (equation 1)

2 u-x+5 y = 13 | (equation 2)

u+x+y+z = 10 | (equation 3)

u-x+y-z = -2 | (equation 4)

Add equation 1 to equation 2:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

x+y+z+u = 10 | (equation 3)

-x+y-z+u = -2 | (equation 4)

Subtract equation 1 from equation 3:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x-y-2 z-3 u = -10 | (equation 3)

-x+y-z+u = -2 | (equation 4)

Multiply equation 3 by -1:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+y+2 z+3 u = 10 | (equation 3)

-x+y-z+u = -2 | (equation 4)

Add equation 1 to equation 4:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+y+2 z+3 u = 10 | (equation 3)

0 x+3 y+2 z+5 u = 18 | (equation 4)

Subtract 1/7 × (equation 2) from equation 3:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+(11 z)/7+(15 u)/7 = 37/7 | (equation 3)

0 x+3 y+2 z+5 u = 18 | (equation 4)

Multiply equation 3 by 7:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+3 y+2 z+5 u = 18 | (equation 4)

Subtract 3/7 × (equation 2) from equation 4:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+(5 z)/7+(17 u)/7 = 27/7 | (equation 4)

Multiply equation 4 by 7:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+5 z+17 u = 27 | (equation 4)

Subtract 5/11 × (equation 3) from equation 4:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+0 z+(112 u)/11 = 112/11 | (equation 4)

Multiply equation 4 by 11/112:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+15 u = 37 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 15 × (equation 4) from equation 3:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+11 z+0 u = 22 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Divide equation 3 by 11:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+3 z+6 u = 33 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 3 × (equation 3) from equation 2:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+0 z+6 u = 27 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 6 × (equation 4) from equation 2:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+7 y+0 z+0 u = 21 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Divide equation 2 by 7:

{x+2 y+3 z+4 u = 20 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 2 × (equation 2) from equation 1:

{x+0 y+3 z+4 u = 14 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 3 × (equation 3) from equation 1:

{x+0 y+0 z+4 u = 8 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Subtract 4 × (equation 4) from equation 1:

{x+0 y+0 z+0 u = 4 | (equation 1)

0 x+y+0 z+0 u = 3 | (equation 2)

0 x+0 y+z+0 u = 2 | (equation 3)

0 x+0 y+0 z+u = 1 | (equation 4)

Collect results:

Answer: |

| {x = 4

y = 3

z = 2

u = 1

Guest Oct 26, 2015

#2**0 **

Thanks Guest,

That was a lot of work !!

Normally this would be done with matrices. I am really rust on matrices I would have to watch a you tube clip or something on how to do it. You answer is great. Did you check that all the numbers made the equations true ? I guess the asker can do that him/herself :))

Melody Oct 27, 2015

#3**+5 **

x+ 2y+ 3z+ 4u=20 (1)

-x + 5 y +2 u=13 (2)

x+ y+ z+ u =10 (3)

-x+y-z+u= -2 (4)

Add (3) and (4) → 2y + 2u = 8 → y + u = 4 → y = 4 - u (5)

Sub this into (2)

-x + 5(4 - u) + 2u = 13 → -x + 20 - 5u + 2u = 13 → -x - 3u = - 7 → x + 3u = 7 → x = 7 - 3u (6)

Subbing into (1) and (4) for x and y we have the following system ;

[7 - 3u] + 2(4 - u] + 3z + 4u = 20

[3u - 7] + [ 4 - u] - z + u = -2

Simplifying both of these, we have

-1u + 3z = 5 (7)

3u - 1z = 1

Multiply the top equation by 3 and add to the bottom, and we have

8z = 16 → z = 2

Subbing this into (7), -1u + 3(2) = 5 → -1u = -1 → u = 1

Subbing this result into (6), we have x = 7 - 3(1) = 4

And subbing u = 1 into (5), we have y = 4 - 1 = 3

So

{x , y, z , u } = { 4, 3, 2, 1 }

CPhill Oct 27, 2015