Arithmetic mean of two unknowns in a system of equations

[travism123]

Homework Statement

When finding the arithmetic mean in a system of equations is there any reason why the method that I am using is wrong?

Find the arithmetic mean of x and y in the following set of equations

Homework Equations

3x + 5y = 65 and
7x + 14y = 175

The Attempt at a Solution

I treated it as a statistics problem where x's and y's are individual units, so...
10x + 19y = 240

mean = 240/29 since mean = total sum/N My answer is 8 and 8/29

I understand that this method might not be the correct way of doing it, but if the number of units is known (10x, and not 10xy) is there any reason that this method of finding the arithmetic mean will not work.

Thank you to anyone who takes the time to look at this.

 

[Mark44]

Homework Statement

When finding the arithmetic mean in a system of equations is there any reason why the method that I am using is wrong?

Yes. See below.

Find the arithmetic mean of x and y in the following set of equations

Homework Equations

3x + 5y = 65 and
7x + 14y = 175

The Attempt at a Solution

I treated it as a statistics problem where x's and y's are individual units, so...
10x + 19y = 240

You can do this, but it is of no help.

mean = 240/29 since mean = total sum/N My answer is 8 and 8/29

This part is not valid. If the left side were 10x + 19x, you could combine these terms to 29x, and it would make sense to divide by 29.

I understand that this method might not be the correct way of doing it, but if the number of units is known (10x, and not 10xy) is there any reason that this method of finding the arithmetic mean will not work.

Thank you to anyone who takes the time to look at this.

The more obvious approach to this problem is to find the solutions x and y to the system of equations, and then find the arithmetic mean of these numbers. This system can be solved pretty easily, giving integer solutions.

 

[travism123]

Thank you Mark.

I guess that I still don't understand why this is wrong. If I solve the equations I get:

x=5 and y=10

So the way that I am looking at it, I have a group of ten 5's and nineteen 10's. If I want to find the mean value for this group then wouldn't it make sense to add them to get my total sum, or 240, and then divide by N, or 29.

x and y are equal between equations, so I am not looking for the mean of x and the mean of y, only the mean of every unknown in the set of equations, or x and y together.

Obviously I would need to solve the equations first if there were an equation with an unknown being multiplied by an unknown like: 10xy + xy + 14y = 392. Then I completely understand why this would not work, but if I know the exact number of x's and y's why can I not simply add the n of each group together to get my N, and then divide the total sum by N?

In the end, isn't it accomplishing the same thing as solving the equations to integers, and then finding the mean.


Just to clarify...I am only looking for the correct answer, and not the correct method. The way that I did it originally seems to be a big shortcut.

Again, thanks very much.

 

[Mark44]

In the end, isn't it accomplishing the same thing as solving the equations to integers, and then finding the mean.

Well, no, because you get different answers. The arith. mean of 5 and 10 is 7.5, which is different from the value you got, which is about 8.28.

What you have said makes some sense, but I have never seen a problem where you are expected to take a system of equations and do anything but find its solution. I guess some context for this problem would be helpful--can you give us the exact wording of the problem?

 

[travism123]

I am positive that you are right Mark.

The actual question was easier:

If a + 2b = 14 and 5a + 4b = 16 what is the average (arithmetic mean) of a and b?

(a) 1.5
(b) 2
(c) 2.5
(d) 3
(e) 3.5

The answer is 2.5.

(6a + 6b = 30)/6 = a + b = 5

5/2 = 2.5

I think that I made it more complicated than necessary. Thanks, and sorry for taking up your time with something so obvious.

 

[Mark44]

No problem. I don't mind helping at all. A favorite saying of mine is, "Keep things as simple as possible, but no simpler."