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Solving A System Of Linear Equations

  • Thread starter Bashyboy
  • Start date
  • #1
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Homework Statement


The problem is to find the values of x and y that solve the following system of linear equations:

6751x + 3249y = 26751

3249x + 6751y = 23249

However, the caveat is that I must contrive a mental process of solving them--I must find x and y in my head.


Homework Equations





The Attempt at a Solution



Rather than waste time, I figured I would analytically solve for x and y, to see if the solved values elucidated some pattern; and then I would contrive a way to solve this problem mentally. I found that x = 3 and y = 2. This didn't seem to be of much advantage.

Before I did this, however, I jotted down a few initial observations:

(1) Notice, the coefficients of the different variables in different equations have the same
value--6751x from the first equation and 6751y from the second equation. So, whatever value is chosen for x must different for y (I am not certain of my reasons of this observation. So, if someone could help me work through this observation, I would appreciate it.).

(2) I also noticed that the RHS of equation 1 is 20000 plus the coefficient of x. A Similar observation can be acquired by looking at equation 2.

I would appreciate someone who could stimulate my thoughts, but please don't just give me the solution. I want to be able to solve this problem on my own, to some degree, of course.
 

Answers and Replies

  • #2
pasmith
Homework Helper
1,737
410

Homework Statement


The problem is to find the values of x and y that solve the following system of linear equations:

6751x + 3249y = 26751

3249x + 6751y = 23249

However, the caveat is that I must contrive a mental process of solving them--I must find x and y in my head.


Homework Equations





The Attempt at a Solution



Rather than waste time, I figured I would analytically solve for x and y, to see if the solved values elucidated some pattern; and then I would contrive a way to solve this problem mentally. I found that x = 3 and y = 2. This didn't seem to be of much advantage.

Before I did this, however, I jotted down a few initial observations:

(1) Notice, the coefficients of the different variables in different equations have the same
value--6751x from the first equation and 6751y from the second equation. So, whatever value is chosen for x must different for y (I am not certain of my reasons of this observation. So, if someone could help me work through this observation, I would appreciate it.).

(2) I also noticed that the RHS of equation 1 is 20000 plus the coefficient of x. A Similar observation can be acquired by looking at equation 2.

I would appreciate someone who could stimulate my thoughts, but please don't just give me the solution. I want to be able to solve this problem on my own, to some degree, of course.
Well, 6 + 3 = 7 + 2 = 5 + 4 = 9 and 1 + 9 = 10, so 6751 + 3249 = 10000. Thus adding the equations gives 10000(x + y) = 50000 so x + y = 5.

You can subtract the second equation from the first to find x - y without having to calculate 6751-3249.
 
  • #3
Mentallic
Homework Helper
3,798
94
If we let

[tex]a=6,751[/tex][tex]b=3,249[/tex][tex]c=20,000[/tex]

Then the equations can be expressed as

[tex]ax+by=c+a[/tex]
[tex]bx+ay=c+b[/tex]

And if you subtract the second equation from the first, you'll get a nice simplification.

(1) Notice, the coefficients of the different variables in different equations have the same
value--6751x from the first equation and 6751y from the second equation. So, whatever value is chosen for x must different for y (I am not certain of my reasons of this observation. So, if someone could help me work through this observation, I would appreciate it.).
That criteria alone isn't sufficient enough to conclude that x and y must be different. For example,

[tex]2x+3y=5[/tex]
[tex]7x+2y=9[/tex]

The coefficient of x in the first equation is equal to the coefficient of y in the second, but the solution to this system is (1,1).

However, if we add the restriction that the coefficient of y in the first is also equal to the coefficient of x in the second as we have in your question, then yes, the solutions to x and y must be different - assuming of course that both equations aren't equivalent.
 
  • #4
1,421
5
A little more prodding towards the solution than I wanted (perhaps you folks should have began with the simple question of what would happen if you added the two equations), but I was able to solve it. Thanks, ya'll.
 

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