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How to solve for three unknown variable that are given in three equations?

Achilles

By Achilles
in Linear Algebra and Group Theory

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Achilles

Achilles

Posted
Posted (edited)

For example:

(1) 3x + 2y + 7z = 100 

(2) 21x + 3y - 5z = 50 

(3) 4x - 2y + 3z = 20 

What is the algebraic method to solve for the sum of unknown variable equaling the number of equations. So if I had 5 unknown variables and 5 equations that included those 5 unknown variables and so on and so on

This was my method using a different equation:

Isolate x from (1)

sub into (2)

Isolate y

Sub y into x

next sub x and y into (3)

Isolate z

And it worked.

Edited by Achilles
wtf

wtf

Posted
Posted

You're basically reinventing linear algebra. You can form a matrix of the coefficients; and if the matrix is invertible, or equivalently has nonzero determinant, you can multiply right side of the equation (the constants) by the inverse of the coefficient matrix to get the answer. The benefit of this method is that it's easily programmed into a computer and requires no insight or cleverness as to what to do first. It's automatic.

The answer to your question about the algebraic method is that you can generalize this by talking about "elementary row operations," and algorithms by which you apply them to reduce your system of equations so that it yields the answer. 

I Googled around and this looks like a pretty good overview of what you're asking about.

https://en.wikipedia.org/wiki/System_of_linear_equations

1
inSe

inSe

Posted
Posted (edited)
  On 9/7/2018 at 5:11 AM, inSe said:

 

fG3MnxG.jpg

 

 

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Ahah! I didn't multiply the 3z by 3 at the end. becomes 99z, 448z + 99z = 547z - 581z = -34z so z=1090/34=32.0588235294

Sub that into the z variable in your isolated y to get y, sub into the y variab in your isolated x variab to get x.

It's tricky to do off instinct when you have this many numbers you can easily mistake

Edited by inSe
-1
studiot

studiot

Posted
Posted
  On 9/7/2018 at 3:30 PM, inSe said:

Okay whoever disliked my solution, explain why it's incorrect or why you think its incorrect

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Nothing to do with me but can you simply list what you think are the values for x, y and z?

inSe

inSe

Posted
Posted
  On 9/7/2018 at 3:48 PM, studiot said:

Nothing to do with me but can you simply list what you think are the values for x, y and z?

Expand  

(x, y, z) = (11.51, -25.25, 16.57)

  On 9/7/2018 at 8:08 AM, inSe said:

ugh Redo!

R7XQlu5.jpg?1

qC7nZik.jpg

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https://www.google.com/search?ei=faGSW7D0O8TLjwSynZbwCQ&q=(3+x+11.509433963)+%2B+(2+x+-25.24528302)+%2B+(7+x+16.566037736)&oq=(3+x+11.509433963)+%2B+(2+x+-25.24528302)+%2B+(7+x+16.566037736)&gs_l=psy-ab.12...49852.113042..114247...2.0..0.105.1712.26j1......0....1..gws-wiz.yiFpSA7iQyE

This has to be right whoever disliked my solution is being a real meany and this hurts my feelings!

Sensei

Sensei

Posted
Posted
  On 9/7/2018 at 4:01 PM, inSe said:

(x, y, z) = (11.51, -25.25, 16.57)

Expand  

Second equation is:

21x + 3y - 5z = 50 

but after using your values we get:
21*11.51 + 3*(-25.25) - 5*16.57 = 83.11

 

-1
inSe

inSe

Posted
Posted
  On 9/7/2018 at 4:08 PM, Sensei said:

Second equation is:

21x + 3y - 5z = 50 

but after using your values we get:
21*11.51 + 3*(-25.25) - 5*16.57 = 83.11

 

Expand  

The second is

 

  On 9/7/2018 at 4:08 PM, Sensei said:

Second equation is:

21x + 3y - 5z = 50 

but after using your values we get:
21*11.51 + 3*(-25.25) - 5*16.57 = 83.11

 

Expand  

if 1 is true

2 is false & 3 is false

Sensei

Sensei

Posted
Posted (edited)
  On 9/7/2018 at 4:15 PM, inSe said:

The second is

 

if 1 is true

2 is false & 3 is false

Expand  

Your values are incorrect. If they would be correct, substitution of x,y,z in equations would give correct results.. That's how you can verify correctness - by substitution (in the all three equations at the same time obviously)..

 

Edited by Sensei
studiot

studiot

Posted
Posted (edited)
  On 9/7/2018 at 4:01 PM, inSe said:

(x, y, z) = (11.51, -25.25, 16.57)

Expand  

So you checked by substituting into the equation(s) as Sensei suggests.

Yes Strange takes the easy way, which funnily enough agrees with my quick method

 

Add equation 3 to equation 1

3x + 2y +7z = 100

subtract

4x - 2y + 3z = 20

yields (eliminates y)

7x         + 3z = 120   ........................4

Subtract 3* equation 1 from 2* equation 2

9x + 6y +21z = 300

42x + 6y -10z = 100

yields (eliminates y) thus I have two equations in x and z only.

-33x + 31z = 200.................................5

Subtract 10 * equation 5 from 31*equation 4

217x + 310z = 3720

-330x + 310z = 2000

yields (eliminates z)

547x = 1720 (exactly)

So x = 1720/547 (exactly)

Which I make 3.1442 as a decimal to 4 decimal places.

 

 

 

 

Edited by studiot
Strange

Strange

Posted
Posted
  On 9/7/2018 at 4:01 PM, inSe said:

hoever disliked my solution is being a real meany and this hurts my feelings

Expand  

Your solutions are wrong (even after posting "Do you know how easy this is for me") and almost illegible so I'm not too surprised. You didn't even remove the old ones when you realised they were wrong.

inSe

inSe

Posted
Posted

Yeah but he asked about the linear algebraic approach which is what mine was. The approach was correct....

  On 9/7/2018 at 4:39 PM, Strange said:

Your solutions are wrong (even after posting "Do you know how easy this is for me") and almost illegible so I'm not too surprised. You didn't even remove the old ones when you realised they were wrong.

Expand  

No they're not, I put a wrong number in there.

 

They were right to begin with

inSe

inSe

Posted
Posted

Okay the algebraic solution this problem turned out to be more multi-faceted than one would intuit but I was clever enough to show the work algebraically which you won't get from these clowns.

 

Let me rewrite it real quick.

-2
inSe

inSe

Posted
Posted (edited)

dqMHBzm.jpg

AwllVDO.jpg

xutkSH8.jpg

Dxa2F4e.jpg

neg all you ye simp it' more accurate than studiot's method

  On 9/7/2018 at 4:45 PM, Strange said:

Just because I wanted to see if it would work!

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neg all you ye simp it' more accurate than studiot's method

you will never be able to invent problem specific solutions on the fly like that

Edited by inSe
-1
inSe

inSe

Posted
Posted
  On 9/8/2018 at 12:11 AM, Sensei said:

You wrote:

50 = -11y -56z + 700

but it should be

50 = -11y - 54z + 700

.......

 

Expand  

Yet it's still more accurate than studiots even when adding it up wrong.

 

intuitive arithmetic

It would have been exact

-3
Sensei

Sensei

Posted
Posted
  On 9/8/2018 at 12:15 AM, inSe said:

Yet it's still more accurate than studiots even when adding it up wrong.

 

intuitive arithmetic

It would have been exact

Expand  

You made error on the first page, in the simple addition and subtraction... Apparently not enough intuitive for you..

 

Strange

Strange

Posted
Posted
  On 9/8/2018 at 12:15 AM, inSe said:

Yet it's still more accurate than studiots even when adding it up wrong.

Expand  

So you think doing it wrong is more accurate than doing it correctly? What planet are you on?

  On 9/8/2018 at 12:15 AM, inSe said:

intuitive arithmetic

Expand  

What is that? You just guess at the results?

  On 9/8/2018 at 12:15 AM, inSe said:

It would have been exact

Expand  

If it hadn't been wrong. Duh.

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