Solve for x and y in the given algebra problem involving fractions

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  • #1
chwala
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Homework Statement:
Solve for ##x## and ##y## in the given problem below;

##\dfrac {x+y}{x-y}##+ ##\dfrac {1}{x+y}##=##\dfrac {5}{25}##
Relevant Equations:
equations
*Kindly note that i created this question (owned by me).

My Approach,

##\dfrac {(x+y)(4x+6y)}{(5x-5y)}##=##-1##

##(x+y)(4x+6y)=-5x+5y##

##\dfrac {4x+6y}{-5x+5y}##=##\dfrac {1}{x+y}##

to get the simultaneous equation,
##4x+6y=1##
##-5x+5y=x+y##
...

##4x+6y=1##
##-6x+4y=0##

giving us ##x=0.076923076##
##y= 0.115384615##

Any positive critic or alternative method is welcome. cheers guys
 
Last edited:

Answers and Replies

  • #2
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Homework Statement:: ##\dfrac {x+y}{x-y}##+ ##\dfrac {x+y}{x-y}##=##\dfrac {5}{25}##
Relevant Equations:: equations


*Kindly note that i created this question.
This is wrong almost from the get-go.
You should at least start from a simplified equation; namely
$$\frac{2x + 2y}{x - y} = \frac 1 5$$
Or better yet, $$\frac{x + y}{x - y} = \frac 1 {10}$$
chwala said:
My Approach,
##\dfrac {(x+y)(4x+6y)}{(5x-5y)}##=##-1##
##(x+y)(4x+6y)=-5x+5y##
##\dfrac {4x+6y}{-5x+5y}##=##\dfrac {1}{x+y}##

to get the simultaneous equation,
##4x+6y=1##
##-5x+5y=x+y##
It doesn't make sense to talk about "a simultaneous equation" when you have only one equation. You will not be able to solve for both x and y when you start with one equation in the variables x and y.
I also have no idea what you did to arrive at the above. Did you make a typo in writing the equation that you created?
chwala said:
...

##4x+6y=1##
##-6x+4y=0##

giving us ##x=0.076923076##
##y= 0.115384615##

Any positive critic or alternative method is welcome. cheers guys
See above.
 
  • #3
chwala
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let me check this again...will get back...yes there is a typo...i have just amended...sorry was bit busy ...
 
  • #4
fresh_42
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It seems someone changed the problem during the discussion.

The current version leads to
\begin{align*}
\dfrac{1}{5}&=\dfrac{x+y}{x-y}+\dfrac{1}{x+y}=\dfrac{(x+y)^2+x-y}{x^2-y^2}\\
x^2-y^2&=5x^2+5y^2+10xy+5x-5y\\
0&=4x^2+10xy+6y^2+5x-5y
\end{align*}
whereas the comments are about a different problem statement.

There are two real roots.
 
  • #5
chwala
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It seems someone changed the problem during the discussion.

The current version leads to
\begin{align*}
\dfrac{1}{5}&=\dfrac{x+y}{x-y}+\dfrac{1}{x+y}=\dfrac{(x+y)^2+x-y}{x^2-y^2}\\
x^2-y^2&=5x^2+5y^2+10xy+5x-5y\\
0&=4x^2+10xy+6y^2+5x-5y
\end{align*}
whereas the comments are about a different problem statement.

There are two real roots.
Yes Fresh, I did amend the question...slight typo error...are my steps correct or you want me to share step by step working?
 
  • #6
fresh_42
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Yes Fresh, I did amend the question...slight typo error...are my steps correct or you want me to share step by step working?
I do not understand your steps.
 
  • #7
chwala
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##\dfrac {x+y}{x-y}##+ ##\dfrac {1}{x+y}##=##\dfrac {5}{25}##

##\dfrac {(x+y)^2}{x-y}##+ ##1##=##\dfrac {x+y}{5}##

##\dfrac {(x+y)^2}{x-y}- \dfrac {x+y}{5}##=##-1##

##\dfrac {5(x+y)^2-(x+y)(x-y)}{5(x-y)}##=##-1##

##\dfrac {(x+y)(5x+5y-x+y)}{5(x-y)}##=##-1##

##\dfrac {(x+y)(4x+6y)}{5x-5y}##=##-1##

##(x+y)(4x+6y)=-5x+5y##

##\dfrac {4x+6y}{-5x+5y}##=##\dfrac {1}{x+y}##

to get the simultaneous equation,
##4x+6y=1##
##-5x+5y=x+y##
...

##4x+6y=1##
##-6x+4y=0##

giving us ##x=0.076923076##
##y= 0.115384615##
 
Last edited:
  • #8
robphy
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##\dfrac {x+y}{x-y}##+ ##\dfrac {1}{x+y}##=##\dfrac {5}{25}##

##\dfrac {(x+y)^2}{x-y}##+ ##1##=##\dfrac {x+y}{5}##

##\dfrac {(x+y)^2}{x-y}- \dfrac {x+y}{5}##=##-1##

##\dfrac {5(x+y)^2-(x+y)(x-y)}{5(x-y)}##=##-1##

##\dfrac {(x+y)(5x+5y-x+y)}{5(x-y)}##=##-1##

##\dfrac {(x+y)(4x+6y)}{5x-5y}##=##-1##

Is the question to
write the given implicit function of x and y
as another implicit function of x and y (possibly in some "standard form")?

You can type (or right-click, Copy to Clipboard as TeX commands) each expression in https://www.desmos.com/calculator to compare
(give different plotting characteristics to each function by long-clicking its control-circle).
 
  • #9
fresh_42
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Fine, now I see how you got to
$$
\dfrac{4x+6y}{-5x+5y}=\dfrac{1}{x+y}
$$
In the next step, you set ##4x+6y=1## and ##-5x+5y=x+y##. This is a nice way to get a solution, and I don't know a better approach. But we can only conclude ##4x+6y=c## and ##-5x+5y=cx+cy## with any ##c\neq 0.##

Now, let me see what your choice of ##c=1## yields:
\begin{align*}
\begin{bmatrix}4&6\\-6&4\end{bmatrix} \cdot \begin{bmatrix}x\\y\end{bmatrix}&=\begin{bmatrix}1\\0\end{bmatrix}\\
&\Longrightarrow \\
\begin{bmatrix}x\\y\end{bmatrix}&=\dfrac{1}{52}\begin{bmatrix}4&-6\\6&4\end{bmatrix}\cdot \begin{bmatrix}1\\0\end{bmatrix}=\dfrac{1}{52}\begin{bmatrix}4\\6\end{bmatrix}
\end{align*}

Are these the numbers you got, and what if ##c\neq 1##?
 
  • #10
chwala
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Fine, now I see how you got to
$$
\dfrac{4x+6y}{-5x+5y}=\dfrac{1}{x+y}
$$
In the next step, you set ##4x+6y=1## and ##-5x+5y=x+y##. This is a nice way to get a solution, and I don't know a better approach. But we can only conclude ##4x+6y=c## and ##-5x+5y=cx+cy## with any ##c\neq 0.##

Now, let me see what your choice of ##c=1## yields:
\begin{align*}
\begin{bmatrix}4&6\\-6&4\end{bmatrix} \cdot \begin{bmatrix}x\\y\end{bmatrix}&=\begin{bmatrix}1\\0\end{bmatrix}\\
&\Longrightarrow \\
\begin{bmatrix}x\\y\end{bmatrix}&=\dfrac{1}{52}\begin{bmatrix}4&-6\\6&4\end{bmatrix}\cdot \begin{bmatrix}1\\0\end{bmatrix}=\dfrac{1}{52}\begin{bmatrix}4\\6\end{bmatrix}
\end{align*}

Are these the numbers you got, and what if ##c\neq 1##?
Yes indeed! its a nice way and it gave me the solution, i used equivalent relationship of fractions i.e ##\dfrac {a}{b}##=##\dfrac {m}{n}## ...to work to solution. Yes fresh, these are the numbers i got;

\begin{align*}
\begin{bmatrix}4&6\\-6&4\end{bmatrix} \cdot \begin{bmatrix}x\\y\end{bmatrix}&=\begin{bmatrix}1\\0\end{bmatrix}\\
&\Longrightarrow \\
\begin{bmatrix}x\\y\end{bmatrix}&=\dfrac{1}{52}\begin{bmatrix}4&-6\\6&4\end{bmatrix}\cdot \begin{bmatrix}1\\0\end{bmatrix}=\dfrac{1}{52}\begin{bmatrix}4\\6\end{bmatrix}
\end{align*}

Bingo!:cool:
 
Last edited:
  • #11
chwala
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I will research later on the case where ##c≠1## and see what comes out, but my thinking is, ...as long as we can equate the fractions then we will always get the solution (by solving the equations simultaneously)...i.e assuming that we are dealing with similar type of problems.
 
  • #13
fresh_42
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...i.e assuming that we are dealing with similar type of problems.
I agree. But you should be aware of the fact that a quotient is always only a relation, i.e. unique up to a factor in numerator and denominator.

Edit: And the solutions do not depend linear on this factor! ##(x,y)=(1/13, 3/26)## is a solution, but as we can see from @robphy's link, ##(x,y)=c\cdot (1/13, 3/26)## are not! A different algebraic transformation in your first steps will lead to a totally different solution. A different value for ##c## will lead to a totally different solution. To solve the problem, you will have to determine pairs ##(x(c),y(c))## that depends (non-linear) on ##c##.
 
Last edited:
  • #14
chwala
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I agree. But you should be aware of the fact that a quotient is always only a relation, i.e. unique up to a factor in numerator and denominator.
Noted mate...cheers.
 
  • #15
SammyS
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Homework Statement:: Solve for ##x## and ##y## in the given problem below;

##\dfrac {x+y}{x-y}##+ ##\dfrac {1}{x+y}##=##\dfrac {5}{25}##
Relevant Equations:: equations


giving us ##x=0.076923076##
##y= 0.115384615##

Any positive critic or alternative method is welcome. cheers guys
Also known as ##(x,y)=\left(\dfrac{1}{13} \ , \ \ \dfrac{3}{26} \right)## .

There are many solutions as pointed out by others.

Two easy to get solutions involve setting ##x## to zero or setting ##y## to zero.

There are at least several integer solutions.

An interesting change of variables is: ##\text{ Let } u = x + y \text{ and } v = x - y \ .##

This gives ##v## as a rational function of ##u## .
 
  • #16
chwala
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Also known as ##(x,y)=\left(\dfrac{1}{13} \ , \ \ \dfrac{3}{26} \right)## .

There are many solutions as pointed out by others.

Two easy to get solutions involve setting ##x## to zero or setting ##y## to zero.

There are at least several integer solutions.

An interesting change of variables is: ##\text{ Let } u = x + y \text{ and } v = x - y \ .##

This gives ##v## as a rational function of ##u## .
True setting ##x=0## or ##y=0## will work but that would be a lazy approach in solving equations...one needs to work from the set problem step by step to realize the solution...yes ofcourse I've noted that we could have other solutions. Thanks Sammy.
 
  • #17
SammyS
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True setting ##x=0## or ##y=0## will work but that would be a lazy approach in solving equations...one needs to work from the set problem step by step to realize the solution...yes of course I've noted that we could have other solutions. Thanks Sammy.
I didn't consider that to be a very significant part of my post.

Altering my suggestion a bit, i found the following substitution to be quite useful.

##\text{ Let } u = x + y \text{ and } v = -x + y \ . ## ( You can solve this pair later for ##x## and ##y##. )

To get you started, you then have

##\displaystyle \frac{u}{-v} + \frac{1}{u} = \frac{1}{5} \ .##

##\displaystyle \frac{u}{-v} = \frac{1}{5} - \frac{1}{u} = \frac{u-5}{5u} ##

## \displaystyle \frac{-v}{u} = \frac{5u}{u-5} ##

And finally: ## \displaystyle v = \frac{-5u^2}{u-5} ##

Among other things, this result leads directly to a parametrization of ##(x,y)## using the single variable, ##u##.
 
  • #19
SammyS
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Possibly useful update:
Wow. Maybe I'll have to spend some time & get familiar with desmos .

I think I have found two asymptotes for this problem. Clearly, for the the graph resulting from the transformation in post #17: ## u = x + y \text{ and } v = -x + y \ .## Taking the ##u## axis as horizontal, the vertical line, ##u=5 ## is a vertical asymptote.
That corresponds to a line ##x+y=5## on @robphy 's desmos graph.
It also, appears to me that the line you attribute to fresh_42, specifically ##4x+6y = c##, is an asymptote for the value, ##c=-25 \ .##
These lines intersect at the point ##\displaystyle (x,y) = \left( \frac{55}{2}, \frac{-45}{2} \right)## .
 
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  • #23
chwala
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Thanks Fresh for the link...i was just looking at the wolframath...
 
  • #25
robphy
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  • Great! Can you please give us a lesson on doing those slider controls - have used Desmos quite a bit but never came across those till now.

(I've been thinking about writing a tutorial... since my recent talks at AAPT meetings have been on using Desmos.)

I’ve been using Desmos since 2015.
The way I learned is to look at what other people have done (“applications” and examples). Then customize for your own use. (There are examples via the hamburger button in the top-left corner of https://www.desmos.com/calculator .)

  • If you type in a mathematical expression with one free variable, it uses that as the independent variable.
  • If you use more than one free variable, it offers you sliders for all of the free variables. You can accept them all, then delete the slider that you want to be the dependent variable.
  • You can get fancier by creating a draggable point (a,b) with two independent variables… or a point on a graph (a, f(a)) where f(x)=cos(x) as an example. The point can play the role of a slider. You can restrict the dragging via the control circle.
You can customize the appearance by clicking the control circle for the expression.

Here's an example: https://www.desmos.com/calculator/zaqc2od8a3 (push "play" of the T-slider)

Other fancy examples:
https://www.desmos.com/calculator/fywfjpb8ug
https://www.desmos.com/calculator/9wtesg4mqg
https://www.desmos.com/calculator/tqxfk6kq9o
https://www.desmos.com/calculator/qalqconaov
https://www.desmos.com/calculator/rhrwohh1zm
https://www.desmos.com/calculator/m1rp8vw6jp
https://www.desmos.com/calculator/awgqxtkqcc
https://www.desmos.com/calculator/8kr9uc9zwu
 
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  • #26
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Thanks very much - will definitely experiment further.
 

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