SOLUTION: SUBSTITUTION AND ELIMINATION 1. 2X+Y-3Z=0 2. -2X+2Y+Z=-7 3. X-4Y-3Z=7 TRYING TO SOLVE FOR X,Y,Z TRIED ELIMINATING X BY ADDING 1&2 WHICH GAVE ME 3Y-2Z=-7 ANS. IS NOW NEW AN

Algebra ->  Coordinate Systems and Linear Equations -> SOLUTION: SUBSTITUTION AND ELIMINATION 1. 2X+Y-3Z=0 2. -2X+2Y+Z=-7 3. X-4Y-3Z=7 TRYING TO SOLVE FOR X,Y,Z TRIED ELIMINATING X BY ADDING 1&2 WHICH GAVE ME 3Y-2Z=-7 ANS. IS NOW NEW AN      Log On


   

Question 121110: SUBSTITUTION AND ELIMINATION
1. 2X+Y-3Z=0
2. -2X+2Y+Z=-7
3. X-4Y-3Z=7
TRYING TO SOLVE FOR X,Y,Z
TRIED ELIMINATING X BY ADDING 1&2 WHICH GAVE ME 3Y-2Z=-7 ANS. IS NOW NEW ANS.4 THEN TRIED TO SOLVE BY ELIMINATING X AGAIN BY USING1&3 WHICH CHANGED NO.3 AFTER MULT. -2(X-4Y-3Z=7 WHICH GAVE ME -2X+4Y+6Z=-14 WHICH IN TURN GAVE ME NEW ANS.5 NOW- THIS IS WHERE I GET CONFUSED DO I SOLVE FOR EITHER Y OR Z IN NEW ANSWERS 4&5 BY FINDING NEW COMMON FACTOR, THEN SOLVING? OR DO I PLACE ANY OF THE ANSWERS INTO ANOTHER PROBLEM?
Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2X+ Y-3Z =0
-2X+2Y+Z =-7
X-4Y-3Z =7
-------------------
Put #3 in the #1 position:
X-4Y-3Z =7
2X+ Y-3Z =0
-2X+2Y+Z =-7
---------------
Subtract twicd #1 from #2; Add twice #1 to #3 to get:
X-4Y-3Z =7
0+9y+3Z =-14
0-6Y-5Z = 7
-----------------
Add #3 to #2 to get:
X-4Y-3Z =7
0+3y-2Z =-7
0-6Y-5Z = 7
----------------
Add twice #2 to #3 to get:
X-4Y-3Z = 7
0+3y-2Z =-7
0+ 0-9Z =-7
----------------
Divide thru #3 by -9 to get:
X-4Y-3Z = 7
0+3y-2Z =-7
0+ 0+Z = 7/9
----------------
Substitute that value of Z into #2 to solve for Y.
Then substitute the Z value and the Y value into #1 to solve for X.
-----------------------
I get :
X = 2.0741
Y = -1.8148
Z = 0.7778= 7/9
=====================
Cheers,
Stan H.




Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
+2x+%2B+y+-+3z+=+0……………….(1)….=> y=+-+2x+%2B+3z ……… (I)

-2x+%2B+2y+%2B+z++=+-7……………..(2)….=> z+=2x+-+2y+-7……………(II)
+x+-+4y+-+3z++=+7………………..(3)

From (I) and (II) we have:
y+=+-2x+%2B+3z…….(I)……….substitute z from (II)
y+=+-2x+%2B+3z…….(
y+=+-2x+%2B+3%282x+-+2y+-7%29…….
y+=+-2x+%2B+6x-+6y+-21%29…….
y+=+4x-+6y+-21%29…….
y+%2B6y+=+4x-21%29…….
7y+=+4x+-+21+
y+=+%284%2F7%29x+-+3……………………(III)



+x+-+4y+-+3z+=+7………………..(3)………..substitute y from (III)

+x+-+4%28%284%2F7%29x+-+3%29+-+3z+=+7………………..substitute z from (II)

+x+-+4%28%284%2F7%29x+-+3%29+-+3%282x+-+2y+-7%29+=+7………………..substitute y from (III)
+x+-+4%28%284%2F7%29x+-+3%29+-+3%282x+-+2%28%284%2F7%29x+-+3%29+-7%29+=+7………………..

+x+-+%28%2816%2F7%29x+-+12%29+-+%286x+-+%2824%2F7%29x+-+18%29+-21%29+=+7………………..solve for x

+x+-%2816%2F7%29x+%2B+12+-+6x+%2B+%2824%2F7%29x+-18+%2B+21+=+7………………..

+-%2816%2F7%29x+-+5x+%2B+%2824%2F7%29x+%2B+15+=+7……………….
++-+5x+%2B+%288%2F7%29x+%2B+15+=+7………………..
++-+5x+%2B+%288%2F7%29x+=+7+-+15………………..
++-+5x+%2B+%288%2F7%29x+=+-+8………………..multiply both sides by 7

++-+35x+%2B+8x+=+-+56………………..

++-+27x+=+-+56………………..

++x+=+-+56%2F-27………………..
x+=+2.07




y+=+%284%2F7%29x+-+3………...................(III)
y+=+%284%2F7%29%28+0.52%29+%96+3+……
y+=+%280.57%29%28+2.07%29+-+3……
y+=+1.1799+-+3
y+=+-1.8201
y+=+-1.8


z+=+2%282.07%29+%96+2%28-1.8%29+-7………… (II)
z+=+4.14+%2B+3.6+-+7+…………
z+=+7.74+-+7…………
z+=+0.74…………

Check:
+2x+%2B+y+-+3z+=+0……………….(1)….=>
+2%282.07%29++-1.8+%96+3+%280.74%29+=+0……………….
+4.14+-+1.8+-+2.22+=+0……………….
+4.14+-+4.02+=+0……………….
+4+-+4+=+0……………….
+0+=+0……………….

Check second and third one too