Show that the ratio ##x+y:x-y## is increased by subtracting ##y##

[RChristenk]

Homework Statement

Show that the ratio ##x+y:x-y## is increased by subtracting ##y## from each term.

Relevant Equations

##x=ky##

##x+y:x-y=\dfrac{x+y}{x-y} \tag1##

Subtract ##y## from each term:

##x:x-2y=\dfrac{x}{x-2y} \tag2##

Assume ##k=\dfrac{x}{y} \Rightarrow x=ky##

##(1)= \dfrac{ky+y}{ky-y}, (2)= \dfrac{ky}{ky-2y}##

Subtract ##(1)## from ##(2)## since we are told by the problem statement ##(2)## is bigger:

##\dfrac{(ky)(ky-y)-(ky+y)(ky-2y)}{(ky-y)(ky-2y)} \Rightarrow \dfrac{k^2y^2-ky^2-(k^2y^2-2ky^2+ky^2-2y^2)}{k^2y^2-2ky^2-ky^2+2y^2} \Rightarrow \dfrac{2y^2}{k^2y^2-3ky^2+2y^2}##

##\Rightarrow \dfrac{2}{k^2-3k+2} \Rightarrow \dfrac{2}{(k-2)(k-1)}##

For ##1<k<2; \dfrac{2}{(k-2)(k-1)}<0## and ##\dfrac{x+y}{x-y}>\dfrac{x}{x-2y}##

For ##k<1## and ##k>2##; ##\dfrac{x+y}{x-y}<\dfrac{x}{x-2y}##

Question: The key to solving this problem was assuming ##k=\dfrac{x}{y} \Rightarrow x=ky##. I know how to plug and chug (obviously), but my question is why is this valid? How does one know ##x## varies proportionally with ##y##? Because ##x## and ##y## could be anything, there's no guarantee they vary proportionally. What are the mathematical rules and assumptions that make this work? Thanks.

 

[PeroK]

Question: The key to solving this problem was assuming ##k=\dfrac{x}{y} \Rightarrow x=ky##. I know how to plug and chug (obviously), but my question is why is this valid? How does one know ##x## varies proportionally with ##y##? Because ##x## and ##y## could be anything, there's no guarantee they vary proportionally.

They don't vary proportionally. The ##k## as defined here is another variable, depending on ##x## and ##y##. Not a constant.

A quicker way is to show that, for any ##x,y## we have$$\frac{x+y}{x-y} \le \frac x {x -2y}$$With equality iff ##y =0##.

PS I'm assuming ##x > 2y \ge 0##.

 

[Mark44]

Homework Statement: Show that the ratio ##x+y:x-y## is increased by subtracting ##y## from each term.
Relevant Equations: ##x=ky##

##x+y:x-y=\dfrac{x+y}{x-y} \tag1##

You wrote ##x = ky## as a relevant equation but it doesn't appear in the problem statement. If this is a given condition, it really should appear in the problem statement.

Something like this:
"Given that ##x = ky##, show that ##\frac{x + y}{x - y} < \frac x{x - 2y}##."

 

[Gavran]

The variable ## k ## is involved because the condition, which must be met, can be expressed using one variable, ## k ##, instead of two, ## x ## and ## y ##, and nothing more. The variable ## k ## could be excluded from the condition and in that case the condition ## k \lt 1 ## or ## k \gt 2 ## would be ## x \lt y ## or ## x \gt 2y ## for ## y \gt 0 ## and ## x \lt 2y ## or ## x \gt y ## for ## y \lt 0 ##.

The problem statement is missing nothing. The problem statement implies that the condition must be included into the result. The problem statement is “Show that the ratio ## x + y : x – y ## is increased by subtracting ## y ## from each term.”, not “## \forall x \in R ## and ## \forall y \in R ## show that the ratio ## x + y : x – y ## is increased by subtracting ## y ## from each term”. In other words values for variables ## x ## and ## y ## are not defined in the problem statement, they must be defined and that is a solution to this problem.

 

[Mark44]

The variable ## k ## is involved because the condition, which must be met, can be expressed using one variable, ## k ##, instead of two, ## x ## and ## y ##, and nothing more.

Since you are not the OP here, this all seems like speculation.

The problem statement is missing nothing. The problem statement implies that the condition must be included into the result.

Again, speculation. The OP did not include the equation x = ky in the problem statement. We should not have to infer what the problem statement includes or doesn't include.