SOLUTION: How do I solve x +3y -6z =7 2x -y +z =1 x +2y + 2z =1 Thanks, Tijuana

Question 33856: How do I solve
x +3y -6z =7
2x -y +z =1
x +2y + 2z =1
Thanks,
Tijuana
Found 2 solutions by venugopalramana, AnlytcPhil:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE THE FOLLOWING AND TRY.IF STILL IN DIFFICULTY PLEASE COME BACK
This is a systems of equations problem and I need to solve for x,y,z. I do not how to get any of the variables or what steps to take to get the answers.
The problem:
2x-5y+ ___=-22
__+y+3z=10
x+___+8z=15
The underscores represents blanks or O
If this had to be an augmented matrix, how would I do that and get those variables?
1 solutions
Answer 20202 by venugopalramana(1462) About Me on 2006-04-16 22:01:33 (Show Source):
2x-5y+ ___=-22...........................I
__+y+3z=10.....................II
x+___+8z=15...........................III
2*EQN.III-EQN.I
2X+16Z-2X+5Y=30+22=52
5Y+16Z=52...........................IV
EQN.IV-5*EQN.II
5Y+16Z-5Y-15Z=52-50=2
Z=2....SUBSTITUTING IN EQN.III
X+8*2=15
X=15-16=-1
SUBSTITUTING FOR Z IN EQN.II
Y+3*2=10
Y=10-6=4
USING MATRIX METHOD...AUGMENTED MATRIX IS
2 -5 0 -22 1 0 0 ?
0 1 3 10 0 1 0 ?
1 0 8 15 0 0 1 ?
NR1=R1/2
1 -2.5 0 -11
0 1 3 10
1 0 8 15
NR3=R3-R1
1 -2.5 0 -11
0 1 3 10
0 2.5 8 26
NR3=R3-2.5*R2
1 -2.5 0 -11
0 1 3 10
0 0 0.5 1
NR3=R3/0.5
1 -2.5 0 -11
0 1 3 10
0 0 1 2
NR2=R2-3R3
1 -2.5 0 -11
0 1 0 4
0 0 1 2
NR1=R1+2.5R2
1 0 0 -1
0 1 0 4
0 0 1 2
HENCE
X=-1
Y=4
Z=2
Answer by AnlytcPhil(1806) (Show Source): You can put this solution on YOUR website!

x + 3y - 6z = 7 2x - y + z = 1 x + 2y + 2z = 1 1. Use the 1st equation to eliminate x from the 2nd equation. 2. Use the 1st equation to eliminate x from the 3rd equation. 3. Use the 2nd equation to eliminate y from the 3rd equation. 4. Solve the 3rd equation for z. 5. Substitute the value of z in the 2nd equation to find y. 6. Substitute the values of y and z in the 1st equation to solve for x. x + 3y - 6z = 7 2x - y + z = 1 x + 2y + 2z = 1 Get rid of the 2x by adding -2 times the 1st equation to 1 times the 2nd: -2[ x + 3y - 6z = 7] 1[2x - y + z = 1] x + 2y + 2z = 1 x + 3y - 6z = 7 -7y + 13z = -13 x + 2y + 2z = 1 Get rid of the x on the bottom left by adding -1 times the 1st equation to 1 times the 3rd: -1[x + 3y - 6z = 7 -7y + 13z = -13 1[x + 2y + 2z = 1 x + 3y - 6z = 7 -7y + 13z = -13 -y + 8z = -6 Get rid of the -y on the by adding -1 times the 2nd equation to 7 times the 3rd: x + 3y - 6z = 7 -1[ -7y + 13z = -13 7[ -y + 8z = -6 x + 3y - 6z = 7 -7y + 13z = -13 43z = -29 Solve the bottom equation for z and get z = -29/43 Replace z by -29/43 in the second equation -7y + 13(-29/43) = -13 Clear of fractions by multiplying by 43 -301y - 377 = -559 -301y = -182 y = -182/(-301) = 26/43 Replace y by 26/43 and z by -29/43 in the 1st equation: x + 3y - 6z = 7 x + 3(26/43) - 6(-29/43) = 7 Clear of fractions by multiplying through by 43 43x + 78 + 174 = 301 43x + 252 = 301 43x = 49 x = 49/43 So x = 49/43, y = 26/43, z = -29/43 Edwin AnlytcPhil@aol.com

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