# Proof that LHS coefficients have to = RHS coefficients

1. Mar 9, 2012

### imiyakawa

1. The problem statement, all variables and given/known data
If;
a*x + b*y = c*x + d*y
x ≠ y
a,b,c,d ≥ 0

Prove that;
a=c
b=d

2. The attempt at a solution
I've been fiddling with this equation and have been getting nowhere.

2. Mar 9, 2012

### Mathoholic!

Try putting that equation in the matricial form. You'll get a 1x2 matrice and a 2x2 matrice on both sides. Since the 1x2 matrices are the same, you can conclude that the 2x2 matrices are also the same, and so a=c and b=d.

3. Mar 9, 2012

### imiyakawa

Hi, thank you for your suggestion. Are you sure that this is a proof? I'm not sure;

[x y] = [x y], hence

[a b] = [c d], hence
a=c,
b=d

I mean it makes complete sense but it's a part of a much longer proof for an assignment so it's important that it's correct

4. Mar 9, 2012

### sjb-2812

Are you sure?

For instance: (8 x 9) + (7 x 4) = (5 x 2) + (30 x 3)

5. Mar 9, 2012

### tiny-tim

hi imiyakawa!
do you mean x is not parallel to y ?

try rewriting it as (a - c)x = (d - b)y

6. Mar 9, 2012

### Ray Vickson

It is not true. For example, the equation 2*x + y = x + 3*y has many solutions, all of the form x = 2*y; the solution satisfies x ≠ y as long as x ≠ 0.

On the other hand, if you really mean that a*x + b*y = c*x + d*y should be true for ALL x ≠ y, then the result is true.

RGV

7. Mar 9, 2012

### Staff: Mentor

The point that Ray Vickson raises is an important one - the difference between a conditional equation (true under certain conditions) and an identity (true for all values of some variable or variables).

For example, the equation 2x + y = 0 has an infinite number of solutions - all of the ordered pairs that satisfy y = -2x.
The identity 2x + y $\equiv$ 0 has only one solution, x = 0, y = 0.

8. Mar 9, 2012

### HallsofIvy

Staff Emeritus
I presume that ax+ by= cx+ dy means that the two sides are equal for all x (given any a, b, c, d, you could find an infinite number of values for x and y which will make that true).

If ax+ by= cx+ dy for all x and y, take x= 1, y= 0. Then take x= 0, y= 1.

9. Mar 9, 2012

### Mathoholic!

Your matricial equation is wrong, you can only multiply matrices like so: (mxn)(nxp); meaning that the first matrix has to have the number of columns equal to the number of lines of the second matrix (n), and the resultant matrix will be (mxp). And so:

ax+by=cx+dy → (1x2).(2x2)=(1x2).(2x2)=(1x2)

The x and y in the 1 by 2 matrices and a,b,c, and d in the 2 by 2 diagonal matrix. If you compute it, you'll have the following:

(ax,by)=(cx,dy) → ax=cx $\wedge$ by=dy

From those equations you conclude that a=c and that b=d.

10. Mar 9, 2012

### Ray Vickson

That conclusion is wrong: the example I gave before (2*x + y = x + 3*y) is a counterexample: this has solutions x ≠ y, but we do NOT have 2 = 1, and/or 1 = 3.

RGV

11. Mar 9, 2012

### Mathoholic!

I've just proved it isn't wrong, how can you say it is? You're misinterpreting the problem, or... perhaps we're tackling it differently.
This equational problem is about the same expression but with different coeficients (alphabetically, at least). Algebraically, I've proved it (for x≠0 and y≠0). For two different expressions:

ax+by=cx+dy

It's obvious that it's not right as a could be a different number then c and the same for b and d. However, in algebra, such is equation is correct because we're only interested to prove it's equal and what it means to be equal (a=c and b=d).
Algebraically, the meaning of the equation is fundamental.
Now, I'm not going to say you're wrong (you aren't, you're only facing it in another point of view, maybe geometrically, with a independent of c and b independent of d, as two separate expressions).
Let me know what you think..

12. Mar 9, 2012

### Staff: Mentor

I disagree. we don't even know what the equation means at this point, and the OP has never cleared this up. IOW, we don't know whether the equation is conditionally true or identically true.

13. Mar 9, 2012

### imiyakawa

Hey everyone. Thanks for so much help on this question.

I need to prove this as part of a broader question in an assignment on utility theory. The equation is this;

aE[u(x)] + (1-a)E[u(y)] = (b+(1-b)a)E[u(x)] + (1-b-(1-b)a)E[u(y)] (1)

Both sides are a gamble, E[u(G(x,y;a))] = E[u(G(x,y;b+(1-b)a))]. I have to give the conditions such that LHS = RHS. Now if you're unfamiliar of a gamble under utility theory see the constraints (below).

(1) is of the form a*x + b*y = c*x + d*y in my original post.

Where;
- each coefficient >= 0
- each coefficient <= 1
- a+b = 1
- c+d = 1
- E[u(x)] ≠ E[u(y)] (there are no other constraints in E[u(x)] and E[u(y)] apart from this one. They can be zero or negative.)

I want to set the coefficients of E[u(x)] equal to each other and the coefficients of E[u(y)] equal to each other to derive required conditions such that LHS = RHS., which is why I was asking for a proof of a*x + b*y = c*x + d*y ... If this is true then I can do what I wanted to do by setting the coefficients equal to each other.

Ray Vickson found an exception for 2*x + y = x + 3*y, where x=2*y given x≠0. Unfortunately I forgot to include the constraint that each coefficient is <= 1 (because we're looking at a formal gamble), a+b=1 and c+d=1 ... I apologise for this, lost my brain when I made the OP.

Last edited: Mar 9, 2012
14. Mar 9, 2012

### imiyakawa

Sorry, x and y are just real numbers. Such as -12.12, 20, 5080.892, etc.

This is not of the form (a)x + (b)y = (c)x + (d)y

Unfortunately I still don't see what's wrong with the matrix that I typed up. Not that you're wrong, I just can't understand what you're saying because my maths education is at quite a low level.

I multiplied a (1*2) matrix with a (2*1) matrix to represent a*x + b*y. Is there something wrong with this?

Last edited: Mar 9, 2012
15. Mar 10, 2012

### Ray Vickson

Whether or not the statement you want to prove is true or false depends on *exactly* what your question meant, which was not obvious from what you wrote. Did you mean to say "for all x ≠ y"? If so, your statement is true and very easy to prove. If you did not mean for all x≠y (but just for some solution with x ≠ y) then the statement is false. Only YOU really know what you meant.

RGV

16. Mar 10, 2012

### imiyakawa

Yes I meant for all x≠y. Thanks for your help.

17. Mar 10, 2012

### Mathoholic!

I'm sorry, I got completely mixed up...there's nothing wrong with that matricial equation.