Question

Use the elimination method to solve the following system of equations.
2x + 4y - 3z = -7
3x + y + 4z = -12
x + 3y + 4z = 4

Answer 1

a) (-6, -2, -1)
b) (6, -2, -1)
c) (-6, 2, -1)
d) (-6, 2, 1)

Answer 2

why not just solve it? :)

Answer 3

pick 2 above, cancel out "z"
pick another 2 from above including the non-included one before, cancel "z"
use the 2 new resultant equations
solve for either "x" or "y"
then extrapolate to the other equations

Answer 4

extrapolate? i dont get how to even set up the question... i dont think i can solve it.

Answer 5

well, pick the 1st two,

2x + 4y - 3z = -7
3x + y + 4z = -12

and cancel "z"

Answer 6

then pick another 2, including the one you didn't use previously, say
1st and 3rd equations

2x + 4y - 3z = -7
x + 3y + 4z = 4

and cancel "z" again

Answer 7

so, you'd get a RESULTANT equation from the 1st cancellation, and another RESULTANT equation from the 2nd cancellation

those 2 RESULTANT equations, will just have "x" and "y", then you'd just solve for either, by eliminating the other

Answer 8

okay ill give it a try

Answer 9

2x + 4y - 3z = -7 times 4
3x + y + 4z = -12 times 3

8x + 16y - 12z = -28
9x + 3y + 12z = -36
-------------------
17x+19y+0 =- 64

Answer 10

2x + 4y - 3z = -7 times 4
x + 3y + 4z = 4 times 3

8x + 16y - 12z = -28
3x + 9y + 12z = 12
--------------------
11x+25y+ 0 = -16

Answer 11

just to let you know...you wouldn't have to cancel out z....you can cancel out any letters x or y also.

Answer 12

and just eliminate from there on those 2, "without a z" equations, and solve for either, x or y

Answer 13

just remember, whatever you cancel out on the first two equations, you have to cancel the same letter out on the last two equations.

Answer 14

indeed

Answer 15

okay i tried them all i think i got answer d) (-6, 2, 1)

Answer 16

2x + 4y - 3z = - 7
3x + y + 4z = - 12 -->(-4)3x + y + 4z = -12
-----------------
2x + 4y - 3z = - 7
-12x - 4y - 16z = 48 (result of multiplying by -4)
-----------------
-10x + 0y - 19z = 41
-10x - 19z = 41

3x + y + 4z = - 12 -->(-3)3x + y + 4z = - 12
x + 3y + 4z = 4
------------------
-9x - 3y - 12z = 36 (result of multiplying by -3)
x + 3y + 4z = 4
-----------------add
-8x + 0 - 8z = 40
-8x -8z = 40

-10x - 19z = 41 -->(4)-10x - 19z = 41
-8x - 8z = 40 -->(-5) -8x - 8z = 40
----------------
-40x - 76z = 164 (result of multiplying by 4)
40x + 40z = - 200 (result of multiplying by -5)
-----------------add
0 - 36z = - 36
-36z = - 36
z = -36/-36 = 1
now sub 1 in for z
-8x - 8z = 40
-8x - 8(1) = 40
-8x -8 = 40
-8x = 40 + 8
-8x = 48
x = - 48/8
x = - 6
now sub in known values in any of the original equations
2x + 4y - 3z = - 7
2(-6) + 4y - 3(1) = - 7
-12 + 4y - 3 = - 7
4y = - 7 + 12 + 3
4y = 8
y = 8/4
y = 2
check....
3x + y + 4z = - 12
3(-6) + 2 + 4(1) = - 12
-18 + 2 + 4 = - 12
-18 + 6 = - 12
-12 = - 12 (correct)

you got it......answer is (-6,2,1)