Systems of Equations?!? i don't get how to do this!?

Question

Solve for X,Y, and Z.

X + 6Y +3Z = 4
2X +Y +2Z = 3
3X - 2Y + Z = 0

I tried substitution, elimination, and matrix... but maybe I'm just doing it wrong?!? ugh! I can't figure it out!

Answer 1

Here's how you can solve it. There are multiple ways, but this is the way I did it using elimination.

For reference, here are the original equations:

x + 6y + 3z = 4
2x + y + 2z = 3
3x - 2y + z = 0

First of all, you need to reduce it to a two-variable, two-equation system. In my method, I eliminate x to start.

To get the first new equation, multiply the first equation by -3 and use the third equation.

-3x -18y - 9z = -12
3x - 2y + z = 0
-----------------------
-20y - 8z = -12

To get the second new equation, multiply the first equation by -2 and use the second equation.

-2x - 12y - 6z = -8
2x + y + 2z = 3
-----------------------
-11y - 4z = -5

These are the two new equations:

-20y - 8z = -12
-11y - 4z = -5

Now, multiply the second new equation by -2 to find the value of y.

-20y - 8z = -12
22y + 8z = 10
----------------------
2y = -2
y = -1

Use substitution to find the value of z.

-11y - 4z = -5
-11(-1) - 4z = -5
11 - 4z = -5
-4z = -16
z = 4

Use substitution in an original equation to find x.

x + 6y + 3z = 4
x + 6(-1) + 3(4) = 4
x - 6 + 12 = 4
x = -2

Therefore, the solution is:

x = -2
y = -1
z = 4

Here's a check to confirm it:

x + 6y + 3z = 4
-2 - 6 + 12 = 4
4 = 4

2x + y + 2z = 3
-4 - 1 + 8 = 3
3 = 3

3x - 2y + z = 0
-6 + 2 + 4 = 0
0 = 0

Answer 2

Eliminate X from first two equations:
-2X - 12Y - 6Z = -8
2X +Y +2Z = 3
add the two:
-11Y -4Z = -5

Eliminate X from last 2 equations:
-6X -3Y -6Z = -9
6X - 4Y + 2Z = 0
add the two:
-7Y -4Z = -9

add those two results together, to eliminate Z:
-11Y -4Z = -5
7Y +4Z = 9
sum:
-4Y = 4
Y = -1

Substitute Y into one of other equations:
7(-1) + 4Z = 9
4Z = 9 + 7 = 16
Z = 4

Then x is easy to find.

With a matrix, they have given you a fairly easy one to do, just eliminate the x value in second row by subtracting twice first one. Eliminate x in third by subtracting 3 times first one. Then subtract multiples of the second row from the first and third one, then subtract multiples of third row from first and second one. This will give the 1 0 0, 0 1 0, 0 0 1 matrix with the answers in right column of course.

Answer 3

Let's subtract multiples of the 1st equation from the other 2, eliminating x:

2x + y + 2z = 3
2x + 12y + 6z = 8
------------------------
........11y + 4z = 5

3x + 18y + 9z = 12
3x - 2y .... + z = 0
------------------------
........20y + 8z = 12

20y + 8z = 12
22y + 8z = 10
------------------
-2y ........ = 2
y = -1

-20 + 8z = 12
8z = 32
z = 4

x + 6(-1) + 3(4) = 4
x = -2

solution (-2,-1,4)

TI-83 verifies it by matrix rref.

Don't feel bad. It is VERY HARD to do one of these without a mistake.

Answer 4

x + 6y + 3z = 4 ..................(1)
2x + y + 2z = 3 ...................(2)
3x - 2y + z = 0 .....................(3)

Now add 2 and 3 and substract 1 from it , we get,
4x -7y = -1 ....................(4)
Multiply (3) by 2 and substract 2 from it , we get,
4x -5y = -3 .....................(5)
Now substract (5) from (4) , we get
- 2y = 2 or y = -1 .Now substitute the value in (4) & x= -2 and similarly putting both the values of x and y in (1) to get the value of z = 4.

So , the answers are x = -2 , y = -1 & z = 4