Find the value of ##5x+5y## in the given equation

[chwala]

Homework Statement

Consider the equation; ##2x+4y##=##\dfrac {1}{2}y##. Find possible values of ##5x+5y##.

Relevant Equations

understanding of equations

This is a problem created by me.

Ok my approach on this,
##2x##=##-\dfrac {7}{2}####y##
##4x=-7y##
##\dfrac {x}{y}=\dfrac {-7}{4}##
##⇒x=-7##, ##y=4##
therefore, ##5x+5y=-35+20=-15##.

there may be more solutions...your thoughts guys.

 

[anuttarasammyak]

There are infinite possible pairs of x and y e.g.
x=-7,y=4
x=-14, y=8
X=1, y=-4/7

Say 5x+5y=A, y=##\frac{A}{5}-x##
By inputting it to 4x=-7y
[tex]4x=-\frac{7A}{5}-7x[/tex]
[tex]A=\frac{55}{7}x[/tex]
As variable x takes any possible value, A does also.

[EDIT]Correction
[tex]4x=-\frac{7A}{5}+7x[/tex]
[tex]A=\frac{15}{7}x[/tex]

 

[chwala]

True, in multiples of ##-7n##...where n lies in R...sorry am typing using my phone ...no latex...##n=1, 2,3, ...## is there a mistake on your last fraction? If you substitute ##x=-7##, we don't seem to arrive at same solution...

 

[fresh_42]

True, in multiples of ##-7n##...where ##n= 2,3, ...## is there a mistake on your last fraction? If you substitute ##x=-7##, we don't seem to arrive at same solution...

These kinds of problems are very badly phrased. You have given a proportion between ##x## and ##y## and try to figure out an absolute value. This cannot be done. It is impossible.

We have ##x=-\dfrac{7}{4}y## so ##5x+5y=-\dfrac{35}{4}y+\dfrac{20}{4}y=-\dfrac{15}{4}y=\dfrac{15}{7}x## and that is all that can be said. There is no way to prefer plugging in a particular value for ##x## or ##y## over any other value. ##x=y=0## would be an easy solution, even without any algebra.

 

[chwala]

Noted, a learning point for me. Thanks Fresh...

just on a side note 'could we then talk of finding the greatest possible value of ##5x+5y##?' ...i can see that we have a straight line function...

 

[fresh_42]

We are given the green line (2x+4y=0.5y) and asked for an intersection with any of the other lines (5x+5y=A). But every non-green line and every parallel to them is valid. So the intersections are all points on the green line, depending on which non-green line we choose, i.e. which value for A. All we have is the slope of the non-green lines, but not where they cross the green line.

 

[Mark44]

This is a problem created by me.
Ok my approach on this,
##2x##=##-\dfrac {7}{2}####y##
##4x=-7y##
##\dfrac {x}{y}=\dfrac {-7}{4}##
##⇒x=-7##, ##y=4##

No, you cannot conclude this without also noting that this is merely one of an infinite number of solutions, including such points (pairs) as (-14, 8), (7, -4) and so on ad infinitum.

The first line of what I've quoted is ##2x = -\frac 7 2 y##.
This is a single equation in two variables, which means that one of the variables can be freely chosen. I've seen you make this over-simplification in at least one of your recent threads.

therefore, ##5x+5y=-35+20=-15##.

No, you cannot conclude this.
Solving for x in your first equation above yields ##x = -\frac 7 4 y##.
Substitute into 5x + 5y:
##5x + 5y = 5(-\frac 7 4 y) + 5y = -\frac{35}4y + \frac{20} 4 y = -\frac{15}4 y##
That's really all you can legitimately say.

The value for y is arbitrary, meaning that 5x + 5y can take on an infinite number of values.

 

[chwala]

What about if we were to find the greatest possible value of ##5x+5y?##

 

[anuttarasammyak]

Referring to A in #2, as
[tex]x[-\infty,+\infty],[/tex]
[tex]A[-\infty,+\infty].[/tex]

[EDIT] ( , ) instead of [ , ]

 

[chwala]

Referring to A in #2, as
[tex]x[-\infty,+\infty],[/tex]
[tex]A[-\infty,+\infty].[/tex]

Thanks will check...we know that ##x## and ##y## are dependant on each other ...but having different sign, one is positive and other negative...I will check on this...thanks Anutta...

 

[chwala]

True, infinity it is...cheers mate.