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

• RChristenk
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.

RChristenk said:
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##.

Last edited:
RChristenk said:
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}##."

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.

Gavran said:
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.
Gavran said:
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.

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