# How many equations

Discussion in 'Math' started by studiot, Apr 23, 2015.

1. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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How many equations would you say this expression represents 1,2, 3 or more?

$\frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} }$

Apr 2, 2009
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OMG !

3. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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Two or more, given the definition of an equation. Or you can write three equations. I see two equations if written in one way. If you add 3 variables, it can be written as five equations. By the way, per the definition of an expression, your example is not one.

4. ### DerStrom8 Well-Known Member

Feb 20, 2011
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An equation is a statement that one thing is equal to another. In the above example you are saying that (x-x1)/(x2-x1) is equal to (y-y1)/(y2-y1) and that (y-y1)/(y2-y1) is equal to (z-z1)/(z2-z1). There's two, and then you are also insinuating that (x-x1)/(x2-x1) is also equal to (z-z1)/(z2-z1). That's three.

5. ### djsfantasi AAC Fanatic!

Apr 11, 2010
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But is the third equation necessary, as it can directly be derived from the first two?

6. ### MrChips Moderator

Oct 2, 2009
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We recognize this as the equations representing a straight line in 3-D space connecting two points, (x1,y1,z1) and (x2,y2,z2).

These represent the orthogonal projections of the line on to the three planes,
x = 0
y = 0
and
z = 0

Each projection in 2-D space is of the form:

y = mx + c

Hence there are three separate equations.

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7. ### DerStrom8 Well-Known Member

Feb 20, 2011
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It is insinuated. It may not be necessary to say because it's already proven, but it is an equation nonetheless.

8. ### WBahn Moderator

Mar 31, 2012
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Are they separate (i.e., independent)?

If one equation is sufficient to define a line in 2D, why does it require two additional, independent equations to define a line in 3D?

If you give me an equation that relates the y-coordinate to the x-coordinate and you give me another equation that relates the z-coordinate to the x-coordinate, what else is needed?

Let's take the equations given in the OP and focus on the two obvious ones:

$\frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{y- y_{1} }{ y_{2}- y_{1} }$

$\frac{y- y_{1} }{ y_{2}- y_{1} }=\frac{z- z_{1} }{ z_{2}- z_{1} }$

It's obvious that I can combine these two equations trivially so as to get the equation

$\frac{x- x_{1} }{ x_{2}- x_{1} } =\frac{z- z_{1} }{ z_{2}- z_{1} }$

And therefore this is not an independent equation. Now, if by "separate" you mean something else, please specify what that is as I'm assuming you mean "independent".

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9. ### WBahn Moderator

Mar 31, 2012
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Since you aren't requiring that the equations be independent, it represents an infinite number of equations.

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10. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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That's an interesting tongue-in-cheek answer.
True of course, but one I never considered anyone would give.

This is really a serious question, not a trick one and I am genuinely interested in people's various reactions and responses to it.
Thus I have tried to be complete in my question, without asking in such a way as to prejudge the issue.

So thank you everyone for your replies so far, keep them coming.

Last edited: Apr 23, 2015
11. ### DerStrom8 Well-Known Member

Feb 20, 2011
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Once again, whether it is required or not, the question was asking how many equations are shown. The answer to that is three.

12. ### wayneh Expert

Sep 9, 2010
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Only 2. The two equal signs provide information about the relationships between x, y and z.
But you'd need another equation to hope to solve for the variables. And going from a=b and b=c to derive a=c, does not give you that 3rd equation.

13. ### WBahn Moderator

Mar 31, 2012
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Depends on what is meant by "solve". Instead of a single (i.e., a point) solution, the solution is an infinite set of points, called a line in this case, any of which satisfy all of the equations in the system.

14. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
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Hmm, now that's a thought. Do the equal signs carry through? If not, you're right--since there's only 2 equal signs, there would be 2 equations. However, if you say x = y = z, does the first sign carry through since the second one is also an equal sign?

Let me know if this doesn't make sense

Matt

15. ### MrChips Moderator

Oct 2, 2009
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If I gave you only two equations of the form

y = mx + c

How would you determine the equation of the line in the third plane?

There appears to me to be three independent equations.

16. ### MrChips Moderator

Oct 2, 2009
14,523
4,282

You can consider this to be one equation that defines the locii of all points falling on the line that intersects two points (x1,y1,z1) and (x2,y2,z2) in 3-D space.

17. ### studiot Thread Starter AAC Fanatic!

Nov 9, 2007
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Yes it is the standard coordinate geometry expression to find the stright line between two points P(x1,y1,z1) and Q(x2,y2,z2) in 3-D space.

The question I am exploring is

Is it one equation or more and if more how many?

18. ### wayneh Expert

Sep 9, 2010
13,638
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My point was only that ƒ(x)=ƒ(y)=ƒ(z) is NOT 3 independent equations, only 2. I arrived at that conclusion from knowing that, in practice, you cannot solve that for all 3 unknowns.

Since one more fact (like x=5) would indeed allow solving for all 3 variables, there must already be two facts, ie. two equations.

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19. ### WBahn Moderator

Mar 31, 2012
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Here is what I think is the interesting point (and maybe the answer will come to me as I type).

If I have

Ax = By + C
and
Dx = Ez + F

It is clear that I can change any of the six coefficients and, in doing so, just change one of the two equations. More specifically, I can alter the relationship imposed by one equation without altering the relationship imposed by the other.

But if I have equations expressed in the form

Gx = Hy + I = Jz + K

Do I still have that ability, or does simply writing the equations in this manner limit what I can and can't do? My gut tells me the two ways of expressing the two relationships should be equivalent in all ways.

I think I see how to show (one way or the other) which it is. I'll doo that in a bit.

20. ### WBahn Moderator

Mar 31, 2012
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I'll agree with you.