Linear system of equations: change in one variable with respect to another

Question

Given a linear system of equations, say with 3 equations in 3 variables $x, y, s$, we can solve for these variables in terms of, say, a constant $c$. Let us assume that $x$ is solved and we get $x = c$.

Thus, $\frac{\Delta x}{\Delta c} = 1$.

But, what if we are asked for $\frac{\Delta x}{\Delta y}$? Does this question make any sense, and if so, how to answer it? To give a simple example: $$s = y -c, \\ y = 2c, \\ x = s$$ What is $\frac{\Delta x}{\Delta y}$?

(This question is asked in an assignment, in a non-mathematical course).

Answer 1

As noted in a comment, it does not really make sense to write $\frac{\Delta x}{\Delta c}$ if $c$ is really a constant. Moreover, if $x$ and $y$ are fully determined by these equations when $c$ is a constant, it does not make sense to write $\frac{\Delta x}{\Delta y}$ either.

This suggests that whoever posed this problem does not really consider $c$ to be a constant. It may be that they interpret $c$ as a "known" parameter in the context of this problem. That is, is it supposed that $c$ will be somehow "given" from outside the problem space, and that we then solve for $x$ and $y$ using this "given" value of $c$.

In that case, we can make both $x$ and $y$ change value by changing the value of $c$.

You can solve the given set of equations assuming $c$ is known, and get $x$ as an expression in $c$ and $y$ as an expression in $c$. Luckily, in this particular problem you should then be able to write $x$ as an expression in $y$, that is, $x = f(y)$, eliminating $c$ from that equation.

I think it will then be fairly easy to see what $\frac{\Delta x}{\Delta y}$ must be. You can still think of the variation of $c$ as the "reason" why $x$ and $y$ are able to vary, but you can find the relative rates at which $x$ and $y$ vary.

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By the way, notice that every term that is added to each side of each equation in your system of equations is either one of the quantities $x, y, s, c$ or a (truly) constant multiple of one of those quantities such as $-c$ or $2c$. As a result, if you multiply every one of the quantities $x, y, s, c$ by the same constant $k$, and substitute the result into the system of equations, you get a set of equations equivalent to the original equations. For example, if we multiply everything by $3$, we get:

\begin{align} 3s &= 3y - 3c\\ 3y &= 2(3c) \\ 3x &= 3s. \end{align}

This implies that if the system of equations has a unique solution for any $c$, all the quantities $x, y, s, c$ will be in the same proportions to each other for any value of $c$; in particular, $x = ry$ for some constant ratio $r$. What you are supposed to do in this problem is to find $r$.

Answer 2

The system here has 3 variables (assuming c is a constant):

$$s = y -c, \\ y = 2c, \\ x = s$$

tells us that:

$$s = c, \\ y = 2c, \\ x = c$$

$\frac{\Delta x}{\Delta c}$, may be a source of confusion. When we say that one of the solutions is $$x=c$$ we don't mean that $$x(c)=c$$ that is $x$ is NOT a function of $c$. We only mean that a specific value of x is equal to $c$, that is: $$x_0=c$$ Accordingly there is no meaning to discuss the change in $x$ w.r.t. $c$. We could say that the above system has a solution vector or a matrix: $$(x_0,y_0,s_0)=(c,2c,c)$$.

The assumption that c is independent of x,y and s is a must for the above to be valid. The problem should specify that up front.