How do you solve the total # of combination for 3 variable for 2 equations

[ultimatebusta]

I've been doing some math contest stuff, and I find a lot of questions that's like 3 different variables, with 2 equations. It asks me to find the number of different possible combinations.

Here's an example:

a+b+c = 110
a+2b+5c = 200

How many different combo of abc are there? (a, b, c are integers. and positive in this question, but not necessarily all questions.)

How do I solve these questions in general?

 

[Mark44]

In general for systems like this, there are an infinite number of solutions. Geometrically, this system of equations represents two planes in three-dimensional space. In this problem, the planes aren't parallel, so the planes intersect in a line. Each point in this line represents a solution to the system.

If a, b, and c are limited to integer values, the solutions will be at lattice points (points all of whose coordinates are integers).

 

[willem2]

If you have only one extra variable, then you can express all the variables as a function of a single variable, and it will be easy to see the conditions on this variable that make all the other variables positive.

 

[arildno]

You can, for example, subtract the first from the second, yielding:

b+4c=90 (3)

Now, subtract (3) from 1, yielding:

a-3c=20 (4)

From (4), we get by rearrangement:

c=(a-20)/3 (5)

Inserting (5) in (3), simplifying and rearranging:
b=-(4/3)a+350/3 (6)

Thus, the combinations (a,b,c) that fulfills the original equation can be written as
(a, (6), (5)), with "a" allowed to have any value.

 

[ultimatebusta]

Well I solved it via

since that 90 - 4c = b. 4c has to be less or equal to 90, which will make c less than 22. Include 0, so there's 23 solutions.

I'm not sure if this would be a way to solve all these types?

 

[willem2]

Well I solved it via

since that 90 - 4c = b. 4c has to be less or equal to 90, which will make c less than 22. Include 0, so there's 23 solutions.

I'm not sure if this would be a way to solve all these types?

All of those with only one extra variable.

If you have n extra variables, you have to determine the number of integer points in an n-dimensional simplex
For example, if you only have a + 2b + 3c = 11 and a>0, b>0 and c>0.

you can write this as a = 11 -2b - 3c, and from this you can see that
2b + 3c < 11, so you need to count the number of integer points in the
triangle b>0, c>0, 2b+3c<11, this will get very hard for higher dimensions,
because the simplex will have so many potential corners.

 

[ultimatebusta]

Yeah I agree. Here's another one of these type of contest question.

a^2 - 4b > 0
a + b = ?

seems very difficult.

Also from all the practise questions I've seen, that the most they'll give you is 4 var with 3 eq. Usually it's 3var with 2 eq.

 

[Mark44]

Here's another one of these type of contest question.

a^2 - 4b > 0
a + b = ?

seems very difficult.

Let's switch variables (letting x = a and y = b) and graph the inequality.

With this change, the inequality becomes x² - 4y > 0, or y < (1/4)x². There are an infinite number of solutions to this inequality -- all of the points below the graph of the parabola y = (1/4)x².

The best you can do for x + y is to write another inequality. You can't say that x + y equals any specific number.