another math question

[Valkyrie144]

here I am again struggling with math.

I have to solve this system.

6x -y +3z =16
x+ 3y -z= -2
3x + 3y -4z = -1

I took the second and third equations and eliminated the x to get my final answer to be -6y -7z =5

Then, I took the first and third equations and eliminated the x to get my final answer to be -7y + 11z = 18

Now, I am to take -6y -7z = 5 and -7y + 11z = 18 and eliminate the z. I end up with -115y = 181 but this can't be right.

Any help would be much appreciated-

 

[Vorpal]

I have to solve this system.

6x -y +3z =16
x+ 3y -z= -2
3x + 3y -4z = -1

I took the second and third equations and eliminated the x to get my final answer to be -6y -7z =5

Then, I took the first and third equations and eliminated the x to get my final answer to be -7y + 11z = 18

You've made an arithmetic error.

Let's skip the variables and look at the coefficients:
+6  -1  +3  = +16
+1  +3  -1  = - 2
+3  +3  -4  = - 1
Subtract 2*(3rd) from the 1st, and 3*(2nd) from the 3rd:
 0  -7  +11 = 18    <-- the -7y + 11y = 18 you have
+1  +3  -1  = -2
 0  -6  -1  = +5    <-- NOT the -6y - 7z = 5 you have

Change the -7 coefficient to -1 and everything works out fine.

y = -1, z = +1

 

[Brian-M]

here I am again struggling with math.

I have to solve this system.

6x -y +3z =16
x+ 3y -z= -2
3x + 3y -4z = -1

Sound like a challenge. I'll give it a try an see if anyone else manages to post the answer before I'm finished.

Let's see. Putting X on one side gets us...

x = (16 - 3z + y) / 6
x = z - 2 - 3y
x = (4z - 1 - 3y) / 3

So now we can remove X to get...

z - 2 - 3y = (16 - 3z + y) / 6
z - 2 - 3y = (4z - 1 - 3y) / 3
(16 - 3z + y) / 6 = (4z - 1 - 3y) / 3

Let's remove the brackets...

6z - 12 - 18y = 16 - 3z + y
3z - 6 - 9y = 4z - 1 - 3y
16 - 3z + y = 8z - 2 - 6y

Now let's get Y all on one side...

y = (9z - 28) / 19
y = (-z - 5) / 6
y = (11z - 18) / 7

So now we can remove Y to get...

(-z - 5) / 6 = (9z - 28) / 19
(-z - 5) / 6 = (11z - 18) / 7
(9z - 28) / 19 = (11z - 18) / 7

Let's remove the brackets...

-z - 5 = z * 54/19 - 168/19
-z - 5 = z * 66/7 - 108/7
9z - 28 = z * 209/7 - 342/7

Now let's try and get Z all on one side...

z + z * 54/19 = 168/19 - 5
z + z * 66/7 = 108/7 - 5
z * 209/7 - 9z = 342/7 - 28

z * (3 + 16 / 19) = 3 + 16/19
z * (10 + 3/7) = 10 + 3/7
z * (20 + 6/7) = 20 + 6/7

z = (3 + 16/19) / (3 + 16 / 19)
z = (10 + 3/7) / (10 + 3/7)
z = (20 + 6/7) / (20 + 6/7)

z = (3 + 16/19) / (3 + 16 / 19)
z = (10 + 3/7) / (10 + 3/7)
z = (20 + 6/7) / (20 + 6/7)

z = 1
z = 1
z = 1

Oh, damn. I'm not sure if I did that right, but let's continue anyway and put it in for Y (from earlier)...

y = (9z - 28) / 19
y = (-z - 5) / 6
y = (11z - 18) / 7

y = - 19 / 19
y = - 6 / 6
y = -7 / 7

y = -1
y = -1
y = -1

That's actually looking a little encouraging. Now let's try for X (from earlier)...

x = (16 - 3z + y) / 6
x = z - 2 - 3y
x = (4z - 1 - 3y) / 3

x = (16 - 3 -1) / 6
x = 1 - 2 + 3
x = (4 - 1 + 3) / 3

x = 2
x = 2
x = 2

Ah, so that's the answer!

X = 2
Y = -1
Z = 1

ETA: I guess Vorpal beat me to it.

 

[Valkyrie144]

Thank you. I'm sure I'll be back with more questions

 

[jt512]

Sound like a challenge. I'll give it a try an see if anyone else manages to post the answer before I'm finished.

Let's see. Putting X on one side gets us...

x = (16 - 3z + y) / 6
x = z - 2 - 3y
x = (4z - 1 - 3y) / 3

So now we can remove X to get...

z - 2 - 3y = (16 - 3z + y) / 6
z - 2 - 3y = (4z - 1 - 3y) / 3
(16 - 3z + y) / 6 = (4z - 1 - 3y) / 3

Let's remove the brackets...

6z - 12 - 18y = 16 - 3z + y
3z - 6 - 9y = 4z - 1 - 3y
16 - 3z + y = 8z - 2 - 6y

Now let's get Y all on one side...

y = (9z - 28) / 19
y = (-z - 5) / 6
y = (11z - 18) / 7

So now we can remove Y to get...

(-z - 5) / 6 = (9z - 28) / 19
(-z - 5) / 6 = (11z - 18) / 7
(9z - 28) / 19 = (11z - 18) / 7

Let's remove the brackets...

-z - 5 = z * 54/19 - 168/19
-z - 5 = z * 66/7 - 108/7
9z - 28 = z * 209/7 - 342/7

Now let's try and get Z all on one side...

z + z * 54/19 = 168/19 - 5
z + z * 66/7 = 108/7 - 5
z * 209/7 - 9z = 342/7 - 28

z * (3 + 16 / 19) = 3 + 16/19
z * (10 + 3/7) = 10 + 3/7
z * (20 + 6/7) = 20 + 6/7

z = (3 + 16/19) / (3 + 16 / 19)
z = (10 + 3/7) / (10 + 3/7)
z = (20 + 6/7) / (20 + 6/7)

z = (3 + 16/19) / (3 + 16 / 19)
z = (10 + 3/7) / (10 + 3/7)
z = (20 + 6/7) / (20 + 6/7)

z = 1
z = 1
z = 1

Oh, damn. I'm not sure if I did that right, but let's continue anyway and put it in for Y (from earlier)...

y = (9z - 28) / 19
y = (-z - 5) / 6
y = (11z - 18) / 7

y = - 19 / 19
y = - 6 / 6
y = -7 / 7

y = -1
y = -1
y = -1

That's actually looking a little encouraging. Now let's try for X (from earlier)...

x = (16 - 3z + y) / 6
x = z - 2 - 3y
x = (4z - 1 - 3y) / 3

x = (16 - 3 -1) / 6
x = 1 - 2 + 3
x = (4 - 1 + 3) / 3

x = 2
x = 2
x = 2

Ah, so that's the answer!

X = 2
Y = -1
Z = 1

ETA: I guess Vorpal beat me to it.

Please stop "helping" with math.

 

[TubbaBlubba]

Brian-M's method is the ideal for these kinds of problems.


(Well, it's actually very useful when dealing with polynomials. First-order algebra, not so much)

 

[Ziggurat]

Brian-M's method is the ideal for these kinds of problems.

Depends on your criteria for "ideal". On the minus side, it's a lot more work than is actually required to solve the problem. But on the plus side, the redundancy acts like a form of error detection/correction. If at any point he got different answers for his x's, y's, or z's, then he'd know that he made a mistake. And since you're usually not likely to make mistakes so as to get the same wrong answer for each of the three lines, this gives added confidence to the result compared to a method which minimizes the number of operations performed.

 

[Brian-M]

I'm not particularly familiar with this kind of problem. To be honest, I had no idea how it was supposed to be solved, which is why I regarded it as an interesting challenge (for me at least). I stuck with basic algebra because that's something I know how to work with.

_{(Yes, I did realize I was being redundant, but I wanted to be on the safe side.)}

 

[Ziggurat]

I'm not particularly familiar with this kind of problem. To be honest, I had no idea how it was supposed to be solved, which is why I regarded it as an interesting challenge (for me at least). I stuck with basic algebra because that's something I know how to work with.

_{(Yes, I did realize I was being redundant, but I wanted to be on the safe side.)}

If you're not comfortable with a problem, then redundancy is a very smart way to keep things safe.

The standard method for solving N equations with N variables is through a sort of recursion. Take one equation, solve for one variable in terms of all the others, and then substitute this newly found expression for that variable in all the other equations. That reduces the problem to N-1 equations with N-1 unknowns (you don't need to keep the first equation around anymore). Repeat to get N-2 equations and N-2 variables, et. until you get down to 1 equation with 1 variable, at which point that variable stops being unknown. Then you backtrack, putting this known variable into one of your equations with 2 variables to solve for a 2nd variable, then those two into the equation with three variables to get a third, etc. until you're back at an equation with N variables, N-1 already known, and solving for the Nth. And then you're done.

This approach isn't actually all that different from what you did, it just drops the redundancy and thereby reduces the overall workload by about half. For a set of linear equations like this, it doesn't even matter which order you do them in, though for more complex equations, the order you solve it in might make the process more or less difficult as a practical matter (though in principle it's all the same).

 

[Ivor the Engineer]

Gaussian elimination.