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**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

5x +2y - z =9

4x - 2y +z =0

I have soved this problem to the point of getting that :

9x = 9

and thus, x = 1.

How do I solve for y and z. Am in College.

*Last edited by MERTICH (2013-12-01 18:12:11)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi MERTICH;

Best you can do is solve it in terms of x or any of the other 2 variables.

Now substituting any value into x will yield a solution to the system.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

Nehushtan wrote:

You made a mistake. It should be

Since there are three variables and only two equations, there will be infinitely many solutions.

I have edited the system. the other z is positive. Sorry guys.

it should be

5x+2y-z = 9

4x-2y+z =0

*Last edited by MERTICH (2013-12-01 18:14:25)*

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**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

bobbym wrote:

Hi MERTICH;

Best you can do is solve it in terms of x or any of the other 2 variables.

Now substituting any value into x will yield a solution to the system.

I am so sorry Bobbym . The other z is positive.

It should be:

5x+2y-z =9

4x-2y+z =0

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi;

No problem, I will just zolve it again.

Substituting for y will all solutions, but unfortunately this post is too small to contain them.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,380

hi MERTICH

Welcome to the forum.

If you had three equations like this, you could make x the subject of one and substitute this into both the other equations.

Then you'd have two equations and two unknowns.

Making y the subject of one and substituting into the other, would enable you to solve for z.

Then you could substitute back to get y and x.

So, as a general rule, three equations with three unknowns leads to a single unique answer for x, y and z. This only applies when the equations are linear .... so not when x^2 etc. Also some sets of three will lead to no solutions or even an infinite number of solutions.

Because you have only two equations with three unknowns, you won't ever be able to get a single answer.

As shown, y and z can be eliminated in a single step by adding the equations together. So x is fixed.

But there's not enough information to fix y and z. As bobbym says, you can choose a value for y and this leads to a value for z. But there are an infinite number of choices.

Is that the whole problem or was this just a part of a problem? Maybe you are now meant to say what this means in the larger problem.

For example, these could be the equations for two intersecting planes. The line x=1 is then the equation for the line of intersection.

Bob

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**MERTICH****Member**- Registered: 2013-12-01
- Posts: 18

Thank you guys, I thought as much, but you know, when doing it alone you doubt yourself. I have learnt something today : a system with three variables and two equations will NOT yield precise values for the variables, except by inserting the infinite number of all possible solutions. Thanx. Will be back soon.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 86,366

Hi;

And welcome to the forum.

Of course that result can be rigorously obtained, but who cares?

Combinatorics is Algebra and Algebra is Combinatorics.

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