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

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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
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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.

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Plug z into x & y

8 hours ago, wtf said:

You're basically reinventing linear algebra.

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1 hour ago, inSe said:

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

ugh Redo!

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

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16 minutes ago, inSe said:

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

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

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11 minutes ago, studiot said:

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

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

7 hours ago, inSe said:

ugh Redo!

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

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4 minutes ago, inSe said:

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

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

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Wolfram Alpha says:

x = 1720/547 = 3.14

y = 6010 / 547 = 10.98

z = 5360 / 547 = 9.80

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2 minutes ago, 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

The second is

6 minutes ago, 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

if 1 is true

2 is false & 3 is false

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9 minutes ago, inSe said:

The second is

if 1 is true

2 is false & 3 is false

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

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36 minutes ago, inSe said:

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

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
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Okay I see where I screwed up at 3 instead of 2y I put 3y

Edited by inSe
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2 minutes ago, inSe said:

Okay I see where I screwed up at 3 instead of 2y I put 3y

Perhaps if you tried it my way you might find it more manageable.

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36 minutes ago, inSe said:

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

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.

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

1 minute ago, 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.

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

They were right to begin with

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11 minutes ago, studiot said:

Yes Strange takes the easy way

Just because I wanted to see if it would work!

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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.

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

7 hours ago, Strange said:

Just because I wanted to see if it would work!

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
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13 minutes ago, inSe said:

You wrote:

50 = -11y -56z + 700

but it should be

50 = -11y - 54z + 700

.......

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2 minutes ago, Sensei said:

You wrote:

50 = -11y -56z + 700

but it should be

50 = -11y - 54z + 700

.......

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

intuitive arithmetic

It would have been exact

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7 minutes ago, inSe said:

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

intuitive arithmetic

It would have been exact

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

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16 hours ago, inSe said:

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

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

16 hours ago, inSe said:

intuitive arithmetic

What is that? You just guess at the results?

16 hours ago, inSe said:

It would have been exact

If it hadn't been wrong. Duh.