x+y+z=2

2x-y+5z= -5

-x-2y+2z=1
asked Oct 26, 2011 in Algebra 1 Answers by anonymous

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2 Answers

x+y+z=2, 2x-y+5z=-5,-x-2y+2z=1

x+y+z=2 <= eq1

2x-y+5z= -5 <= eq2

-x-2y+2z=1
<= eq3
 
eq1 + eq2
  x + y + z = 2 add to  2x - y + 5z = - 5
 
 3x + 6z = - 3
 x + 2z = -1 <= eq4
 
eq1 + eq3
   x + y + z = 2 add to  - x - 2y + 2z = 1
 y + 3z = 3 <= eq5
 
in eq4 : x = - 1 - 2z <= 6
 
eq6 subtitute to eq3
 
- x -2y+2z = 1 <= eq3
-(-1 - 2z) - 2y + 2z = 1
1 + 2z - 2y + 2z = 1
- 2y + 4z = 0
4z = 2y
2z = y 
 
z = y/2 < eq7
 
eq7 in eq5
 y + 3z = 3 <= eq5
 y + 3(y/2) = 3
2y + 3y = 3(2)
5y = 6
 
y = 6/5
 
y in eq2
2x-y+5z= -5 <= eq2
2x - (6/5) + 5z = - 5
2x - (6/5) + 5z + 5 = 0
10x - 6 + 5(5)z + 5(5) = 0
10x + 25z +25  - 6 = 0
10x + 25z = -19 <eq8
 
x + 2z = -1  <= eq4
x = - 2z - 1 <= x into eq8
 
10(-2z - 1) + 25z = -19
- 20z + 10 + 25z = -19
+ 10 + 5z = -19
5z = - 19 - 10
5z = - 29
 
z = -29/5
 
What we have computed  y = 6/5; z = -29/5 enter into eq1
 
x+y+z=2 <= eq1
x + (6/5) + (-29/5) = 2
x + (6/5) + (-29/5) - 2  = 0
[(5)x + 6 - 29 -(5)2] = 0
5x + 6 - 29 - 10 = 0
5x - 33 = 0
5x = 33
 
x = 33/5
 
Summary:   x = 33/5 : y = 6/5: z = -29/5 ◄ Ans
 
answered Oct 31, 2011 by E♥Lim Level 3 User (2,700 points)
 

x+y+z=2, 2x-y+5z=-5,-x-2y+2z=1

x+y+z=2 <= eq1

2x-y+5z= -5 <= eq2

-x-2y+2z=1
<= eq3
 
eq1 + eq2
  x + y + z = 2 add to  2x - y + 5z = - 5
 
 3x + 6z = - 3
 x + 2z = -1 <= eq4
 
eq1 + eq3
   x + y + z = 2 add to  - x - 2y + 2z = 1
 y + 3z = 3 <= eq5
 
in eq4 : x = - 1 - 2z <= 6
 
eq6 subtitute to eq3
 
- x -2y+2z = 1 <= eq3
-(-1 - 2z) - 2y + 2z = 1
1 + 2z - 2y + 2z = 1
- 2y + 4z = 0
4z = 2y
2z = y 
 
z = y/2 < eq7
 
eq7 in eq5
 y + 3z = 3 <= eq5
 y + 3(y/2) = 3
2y + 3y = 3(2)
5y = 6
 
y = 6/5
 
y in eq2
2x-y+5z= -5 <= eq2
2x - (6/5) + 5z = - 5
2x - (6/5) + 5z + 5 = 0
10x - 6 + 5(5)z + 5(5) = 0
10x + 25z +25  - 6 = 0
10x + 25z = -19 <eq8
 
x + 2z = -1  <= eq4
x = - 2z - 1 <= x into eq8
 
10(-2z - 1) + 25z = -19
 
correction
 
- 20z - 10 + 25z = -19
- 10 + 5z = -19
5z = - 19 + 10
5z = - 9
 
z = - 9/5
 
What we have computed  y = 6/5; z = -29/5 enter into eq1
 
x+y+z=2 <= eq1
x + (6/5) + (-9/5) = 2
x + (6/5) + (-9/5) - 2  = 0
x + 6/5 - 9/5  - 2  = 0
[(5)x + 6  - 9  - (5)2] = 0
 
5x + 6 - 9 - 10 = 0
5x - 3 = 0
5x = 3
 
x = 3/5
 
correction made
 
x = 3/5 :  y = 6/5 :  z = -9/5  ◄ Ans
answered Oct 31, 2011 by E♥Lim Level 3 User (2,700 points)
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