# Is it possible to solve this?

1. May 27, 2014

### alikthename

Hi everybody.

Is it possible to solve this equation?

a+b+c = 59;
d+e+c = 52;
f+b+g = 95;
a+h+i = 84;
d+h+g = 75;
f+e+i = 97;

2. May 27, 2014

### MisterX

What you have there is a system of linear equations.

Whether such systems have a solution is a well known problem. Students learn general methods for studying and solving such systems in their first linear algebra course.

Notice that you have 9 variables but only 6 equations. This means either there are no solutions, or there are infinitely many solutions.

To answer your question, you can try solving it. One way to do this would be to form an augmented matrix, and then use a method to get the matrix into reduced row echelon form. Many graphing calculators can do this for you, as well as other computer programs. It is also possible to do it by hand.

3. May 27, 2014

### Joffan

5 equations in 9 variables... you'll get rather a lot of free variables left in your solution. I note that the sixth equation is eliminated as (1)+(5)+(6) is the same as (2)+(3)+(4).

4. May 27, 2014

### alikthename

Thank you very much for answers.
Here is another point. Even if I make add up to these 6 equations another 3 and make it 9 equations like
a+b+c = 59 ;
d+e+c = 52 ;
f+b+g = 95 ;
a+h+i = 84 ;
d+h+g = 75 ;
f+e+i = 97 ;
f+c+d = 32 ;
a+h+i = 84 ;
g+e+b = 115;

Here is the result I get:
1 1 1 0
0 0 0 1
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

Which if I get it right means "no solution"

Is that because I have 1's as coefficients? Or because I have 0's as coefficients for other variables?

5. May 27, 2014

### Staff: Mentor

No, for example

a+b=1
b+c=2
a+c=3

has a nice solution of a=1, b=0, c=2 (despite using 1's and 0's for coefficients).

It is just that not every set of linear equations has a solution. This is not something for a short post on the forum. You may want to google for "system of linear equations", there are many pages that discuss properties of such systems and methods used to solving them.

6. May 27, 2014

### HallsofIvy

Staff Emeritus
It's hard to tell if you have it right because it is impossible to tell what you have done. For one thing you haven't said what your "result" means- you have not said what those "1"s and "0" in your array mean. I suspect that you set this system up as and augmented matrix and then row reduced but, because you have 9 unknowns in your equations, you should have 10 columns, one for each unknown and one for the right side.
Taking your three unknowns for this system to be "a", "b", and "c", those equations become
a+ b+ c 0 and 0= 1. Since that last equation is not true for any a, b, or c, there is no solution.

Last edited: May 27, 2014
7. May 27, 2014

### Staff: Mentor

Actually you still don't have 9 equations.

8. May 27, 2014

### Joffan

Your set of 9 equations - which as Borek says has a direct duplicate, never mind that for now - is encoded as a matrix equation like this:

$$\begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 0 & 0 & 1 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 1\\ 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0\\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1\\ 0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 & 0\\ \end{pmatrix} \begin{pmatrix} a\\b\\c\\d\\e\\f\\g\\h\\i\\ \end{pmatrix} = \begin{pmatrix} 59\\52\\95\\84\\75\\97\\32\\84\\115\\ \end{pmatrix}$$

Start from there if you want to make any sense of what you have.

9. May 28, 2014

### Joffan

Just to demonstrate that there are indeed solutions to your 9 equations:

a = 20 (arbitrary choice)
b = 20 (arbitrary choice)
c = 19
d = 5 (arbitrary choice)
e = 28
f = 8
g = 67
h = 3
i = 61

Due to the relationships between the equations, I can make any choices for a, b and d (for example) and still find solutions for the other six variables.