[Guest]

Hello Thalia and other homework helpers,

I am given this "system" of equations as follows:

(1) c + s + r = 100

(2) 50c + 10s + .5r = 1,000

Question: Find the solution of c, s, and r in integers.

So how does one solve two equations with 3 variables? I have seen one equation with one variable, and two equations with 2 variables, but never the two equations with 3 variables. I guess to solve this problem then is to get rid of one of the variable. It looks like I can easily get rid of the variable r because if I can manage to convert the .5r in equation 2 to just r, then I can subtract the r in equation 1 from it. So this is how I would approach it because my goal is to solve the 2 equations with just two variables.

So if I multiply the equation 2 by 2, I will get 2(50c + 10s + .5r = 1,000) or ===> 100c + 20s + r = 2000

If I subtract equation 1 from this equation 2, then I will get...

100c + 20s + r = 2000

-(c + s + r = 100)

=======================

99c + 19s = 1900

This is where I am stuck. Can the values of c and s be found methodically? At this point I am using the trial and error strategy and it does not take me anywhere.

Edited April 29, 2011 by NuttyNumbers

[curr3nt]

Since c, s and r are integers you know the following.

From #2

0 <= c < 20 since 50c < 1000

Also

10(c + s + r) = 1000 => 50c + 10s + r/2 = 10c + 10s + 10r

From that I can get the values of c and r which leads to s.

[Guest]

Since c, s and r are integers you know the following.

From #2 0 <= c < 20 since 50c < 1000 Also 10(c + s + r) = 1000 => 50c + 10s + r/2 = 10c + 10s + 10r From that I can get the values of c and r which leads to s.

Thank you, Curr3nt, for your suggested steps to the solution. Before reading your comments, I would have never thought that "c" is at least equal to 0 or less than 20 at all. Your ability to consider different angles surrounding the two equations is simply amazing to me.

Edited April 29, 2011 by NuttyNumbers

[unreality]

Yes the "extra" piece of info here is that they have to be integers. This doesn't always guarantee a unique solution, in fact these types of problems are generally very difficult (see Diophantine equations or Fermat's last theorem) but in your case, that was the extra info needed to make up for the lack of a 3rd equation.

The problem with curr3nt's assertion however is that he assumes they are positive integers when integer doesn't specify sign.

(1) c + s + r = 100

(2) 50c + 10s + .5r = 1,000

Question: Find the solution of c, s, and r in integers.

take the original equation set and multiply the first by 20 and the second by 2 to get:

20c + 20s + 20r = 2000

100c + 20s + r = 2000

Once again, subtract the first from the second:

80c - 19r = 0

80c = 19r

Here is where the integer thing helps. 19 is prime, thus 80 and 19 are coprime. It basically means that c has to be a multiple of 19 and r has to be a multiple of 80. This can be more easily seen by dividing them out:

c = 19r / 80

r clearly needs to be a multiple of 80 to make c an integer, since 19 never divides 80. Therefore r = 80k, where k is some integer (including zero and negative integers), and thus c = 19k

Anyway, going back to the original equation:

s+r+c=100

s + 99k = 100

So now we have all three as a function of a single parameter:

r = 80k

c = 19k

s = 100 - 99k

If they have to be positive, clearly the only solution is from the parameter k=1 (leading to 80,19,1) but it's important to note that ANY integer choice for k will satisfy this. You can check this yourself. It works for k=0, k=2, k = -1, etc. But only k=1 gives positive integer answers. So really it depends on what it's asking for

[Guest]

Unreality, thanks for your wonderful help. After reading your explanation, I was completely speechless and in awe of your indepth understanding. You know your mathematical art really well. Yeah to you!