How can I solve this tricky system of equations?

[zeronem]

I found this System of Equations to be a real pain, I've come up with a lot various answers and some of them have worked, but then when i try it on others, they don't work.

x + 2y = 5
3x +4z = 2
2y + 3w = -2
3z - 2w = 1

Ive tried solving this so many times. I've gotten z = 29/35, x= 323/35, y = -74/35, w = 26/35. They work on some, but then it doesn't work on the x. I've also gotten w = 17/35, x = -22/105, z = 23/35, and y = 547/210. Evidently I'm making some mistakes in some areas of solving this system. Any mistake can cause huge catastrophe in Math. Please help me out here.

 

[skeeter]

x + 2y = 5
3x +4z = 2
2y + 3w = -2
3z - 2w = 1

from the 1st equation, x = 5-2y
sub into 2nd equation ...
3(5-2y) + 4z = 2

from 4th equation, w = (3z-1)/2
sub into 3rd equation...
2y + 3[(3z-1)/2] = -2

now you have two equations with y and z as variables...
3(5-2y) + 4z = 2 ... 2y + 3[(3z-1)/2] = -2
15 - 6y + 4z = 2 ... 4y + 3(3z-1) = -4
-6y + 4z = -13 ... 4y + 9z - 3 = -4
... 4y + 9z = -1

4(-6y + 4z = -13)
6(4y + 9z = -1)

-24y + 16z = -52
24y + 54z = -6
-----------------
70z = -58
z = -58/70 = -29/35

3z - 2w = 1
-87/35 - 2w = 1
w = -61/35

3x + 4z = 2
3x - 116/35 = 2
x = 62/35

x + 2y = 5
62/35 + 2y = 5
2y = 5 - 62/35
y = 113/70

The person that made up this system was definitely a numerical sadist...some ugly solutions, but they all work.

 

[zeronem]

Thanks man, The Frustration has left. I might have solved it correctly but my work on the paper was everywhere and unorganized. You brought it out clearly with a method I was familiar with, only i did not use it hehe. Once again thank you.

 

[BoulderHead]

Zeronem,
What method were you using?

Here’s one;
If you were allowed to use matrix operations to solve this (or want another way to check your answer with a calculator) you could rewrite it like this;

x + 2y = 5
3x +4z = 2
2y + 3w = -2
3z - 2w = 1

becomes;

0W + 1X + 2Y + 0Z = 5
0W + 3X + 0Y + 4Z = 2
3W + 0X + 2Y + 0Z = -2
-2W + 0X + 0Y + 3Z = 1

Which becomes;

0 1 2 0 * W = 5
0 3 0 4 * X = 2
3 0 2 0 * Y = -1
-2 0 0 3 * Z = 1

(sorry, I don't know how to put brackets into that)
Which is in the form of;

Ax = y, or if you prefer; Ax = C


...Where you can pre multiply both sides by the inverse of matrix A to solve for column vector x (the one with the W, X, Y, and Z);
x = A^-1y

This would give the same answers Skeeter showed.
While it may look rather intimidating if you haven't learned about matrix operations, it is conceptually quite easy, just tedious in practice. However, for a modern calculator it is very fast and straightforward.

 

[zeronem]

Yes, i actually used to know the Augmented Matrix Method but I forgot it in time.
Here was the method I was useing.

x + 2y = 5
3x + 4z = 2
2y + 3w = -2
3z - 2w = 1

-3(x + 2y)=5(-3)---> -3x - 6y = -15 the 3 and -3 cancel.
3x + 4z = 2------> 3x + 4z = 2
= -6y + 4z = -13

Then i would take
2y + 3w = -2
3z - 2w = 1
make the w's cancel each other
2(2y + 3w) = -2(2) ------> 4y + 6w = -4
3(3z - 2w) = 1(3)--------> 9z - 6w = 3
= 4y + 9z = -1
.
Then we take -6y + 4z = -13 and 4y + 9z = -1
and cancel the y
2(-6y + 4z) = -13(2) -----------> -12y + 8z = -26
3(4y + 9z) = -1(3) -------------> 12y + 27z = -3
35z = -29
z = -29/35
This was the method I was using. I think I could of pulled it off if my work wasn't so unorganized on the paper I was using. I usually get completely careless in these problems when the Teacher teaches us stuff that is like kindergarden work for me.