# Proof that LHS coefficients have to = RHS coefficients

• imiyakawa
Thanks for saying that I am not wrong, however, it is not a matter of opinion: I have presented a counterexample to the OP's "proof." That counterexample shows that the conclusion (a=c and b=d) is incorrect, and that the OP's "proof" is also incorrect. The OP has not yet responded to my counterexample, but I think it is clear that the OP's "proof" is not correct.RGVI am sorry, I did not mean to say that you are wrong. I meant to say that we are looking at it in two different ways. I wasn't trying to invalidate your point of view by any means.
imiyakawa

## Homework Statement

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.

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.

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=dI 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

imiyakawa said:

## Homework Statement

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.

Are you sure?

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

hi imiyakawa!
imiyakawa said:
x ≠ y

do you mean x is not parallel to y ?

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

imiyakawa said:

## Homework Statement

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.

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

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.

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.

imiyakawa said:
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=dI 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

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.

Mathoholic! said:
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.

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

Ray Vickson said:
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

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..

Mathoholic! said:
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.
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.
Mathoholic! said:
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..

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.

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tiny-tim said:
do you mean x is not parallel to y ?

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

sjb-2812 said:
For instance: (8 x 9) + (7 x 4) = (5 x 2) + (30 x 3)

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

Mathoholic! said:
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.

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:
imiyakawa said:
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?

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

Ray Vickson said:
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

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

imiyakawa said:
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?

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

## 1. What does LHS and RHS stand for in this context?

In mathematics, LHS stands for "Left Hand Side" and RHS stands for "Right Hand Side". These terms refer to the two sides of an equation, with the equal sign in between.

## 2. Why do the coefficients on the left and right sides of an equation have to be equal?

The coefficients on the left and right sides of an equation have to be equal because of the fundamental property of equality in mathematics. This means that whatever operation is performed on one side of the equation must also be performed on the other side in order to maintain the balance and equality of the equation.

## 3. Can the LHS and RHS coefficients ever be different?

No, the LHS and RHS coefficients cannot be different. If they were, it would result in an unbalanced equation and would not be a true statement.

## 4. Are there any exceptions to the rule that LHS coefficients have to equal RHS coefficients?

There are no exceptions to this rule. It applies to all equations, regardless of the variables or operations involved.

## 5. How does this concept relate to the Law of Conservation of Mass in chemistry?

In chemistry, the Law of Conservation of Mass states that matter cannot be created or destroyed in a chemical reaction. This is similar to the concept of LHS and RHS coefficients being equal in an equation. It ensures that the amount of matter on both sides of the reaction remains balanced and constant.

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