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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

Best Answer 

 #1
avatar
+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
avatar
+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
avatar+93691 
0

Thanks Guest,

That was a lot of work  !!      smiley

 

 

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
avatar+90088 
+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 }

 

 

 

cool cool cool

CPhill  Oct 27, 2015

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