Solving A System Of Linear Equations

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The discussion revolves around solving a system of linear equations: 6751x + 3249y = 26751 and 3249x + 6751y = 23249, with an emphasis on finding a mental process for the solution. The initial analytical solution yielded x = 3 and y = 2, but the poster sought deeper insights into the relationships between the coefficients. Observations noted that the coefficients in both equations are symmetric, suggesting that x and y must differ, but this was debated. Ultimately, a method was proposed to simplify the equations by adding and subtracting them, leading to a clearer path to the solution. The discussion highlights the importance of understanding the structure of the equations to facilitate mental calculations.
Bashyboy
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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.
 
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Bashyboy said:

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.
 
If we let

a=6,751b=3,249c=20,000

Then the equations can be expressed as

ax+by=c+a
bx+ay=c+b

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

Bashyboy said:
(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,

2x+3y=5
7x+2y=9

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.
 
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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