I have a 3 variable problem, but only 2 equations, how would I solve it?

Question

The 2 equations are
a + b + c = 100
10a + 3b + .5c = 100

so I guess you could also say that
a + b + c = 10a + 3b + .5c

Solve for a, b & c .
(The answer is 5, 1 & 94 but I would like to know how to solve it mathematically.) Thanks.

Note: Sorry, I forgot to mention that the answers had to be positive integers.

Note: It probably would have been clearer if I had written .5c as 1/2c or c/2, I think some people missed the decimal.

Answer 1

If given no other information, this equation has an infinite number of solutions, of which that is only one of them. So I assume that the context of the problem was one where you are supposed to find only the solutions where a, b, and c are positive integers. In this case, you use substitution in one equation, and then set up a congruence relation in the other. I will assume (perhaps incorrectly) that you are familiar with modular arithmetic.

In the first equation, solve for a:

a=100-b-c

Substitute into the second equation:

10(100-b-c) + 3b + c/2 = 100

Expand:

1000 - 7b - 19c/2 = 100

Here we choose to solve for c. Add 19c/2-100:

900-7b = 19c/2

Divide by 19/2:

(1800-14b)/19 = c

Now, for c to be an integer, 1800-14b must be evenly divisible by 19. So we set up a congruence:

1800-14b≡0 mod 19

We know that 1800 ≡ 14 mod 19, so this means

14-14b≡0 mod 19

Adding 14b to both sides and dividing by 14, we see that:

b≡1 mod 19

Which by definition means that b=1+19k for some integer k. We also know that k cannot be negative, since that would make b negative. Substituting this into the formula for c, we find that:

c = (1800-14(1+19k))/19 = 94-14k.

Finally, substituting this into the first equation, we see that:

a = 100 - b - c = 100 - (1+19k) - (94-14k) = 5 - 5k

Since a is positive, this means that k cannot be positive either, since that would make a≤0. So in fact, k must be zero, leaving the unique solution of a=5, b=1, and c=94.

Answer 2

As has been mentioned, you can't solve this (with one single answer) if you're allowing a, b, and c to be real numbers. Also there isn't just one single answer if you let them be any integers. However, I think the context here is that a, b, and c are integers between say 0 and 100.

The solution (5,1,91) does work though: the second equation has 0.5, not 5

These sort of equations aren't typically easy, but this one doesn't seem too bad. Off the top of my head, I would say we should use the first two equations and eliminate one variable, I'll choose c. Then we get

19a + 5b = 100
b = 20 - 19a/5

So we can see that if b is to be an integer, a must be divisible by 5. Now if we are only allowing solutions between 0 and 100, then we can also deduce from this latest equation that a/5<2, since if a/5>=2, we would come up with a negative answer for b. Therefore a=0 or 5. That's quite a narrow set, so let's just try them both. If a=0, then b=20, and c=80. If a=5, then b=1 and c=94, as you have.

Considering that I came up with two solutions, perhaps the contextual restriction is that the numbers be integers strictly between 0 and 100. Actually, I never did use the restriction that the solutions be less than 100. So really the restriction could be positive integers (since positive does not include 0).

But the important thing to remember is that this has infinitely many solutions without the restrictions we placed on the solution(s). In general for real number equations, we need at least as many non-trivial equations as variables.

Answer 3

The non-modular way -

(1) a + b + c = 100
(2) 10a + 3b + c / 2 = 100

From (1), c = 100 - a - b

Substituting into (2) gives:
10a + 3b + (100 - a - b) / 2 = 100

Multiplying through by 2 gives:
20a + 6b + 100 - a - b = 200

or, 19a + 5b = 100

Now find b (variable with smallest constant):
b = (100 - 19a) / 5 = 20 - 19a / 5

Take out the largest number
of integers from 19a / 5:
b = 20 - 3a - 4a / 5

Each term must be an integer, so let 4a / 5 = J.
Therefore, a = 5J / 4

Again, take out largest number of integers:
a = J + J / 4

All must be integers again, so let J / 4 = K.
Therefore, J = 4K.

Now we are rid of fractions, substitute back:
J = 4K implies that a = 4K + 4K / 4 = 5K.

a = 5K implies b = 20 - 3(5K) - 4(5K)/5 = 20 - 19K.

Which implies c = 100 - 5K - (20-19K) = 80 + 14K.

So we have: a = 5K, b = 20 - 19K, c = 80 + 14K.

Now, K must be >= 1, otherwise 'a' will be <= 0.
Also, K must be <= 1, otherwise 'c' will be > 100.
The only solution then, is that K = 1.
Substituting then gives: a = 5, b = 1, c = 94.

Answer 4

To get rid of the fraction, multiply the second equation by 2. So we have
a + b + c = 100
20a + 6b + c = 200

From the first equation get c = 100 - a - b, and substitute into the second equation, getting
20a + 6b + (100 - a - b) = 200, or
19a + 5b = 100

The strategy for solving this kind of equation in integers is to find one solution, and then all others can be calculated by adjusting the original solution. In this case, we can easily guess a solution, namely a = 0 and b = 20.

Note that 19*5 = 95. If we increase 19a by 95 and decrease 5b by 95, then we will still have a solution. But increasing 19a by 95 means increasing a by 5. And decreasing 5b by 95 means decreasing b by 19. Thus we can get infinitely many other solutions:
a, b
0, 20
5, 1
10, -18
15, -37
etc.

Or, we could go the other way:
a, b
0, 20
-5, 39
-10, 58
etc.

Of course, with each a,b solution, you can get the c to go with them from the equation c = 100 - a - b.

The general solution is
a = 5t
b = 20 - 19t
c = 80 +14t
where t is any integer.
You can check the solution by simple substitution.
It's easy to see that the only value for t that make a,b,c all positive is t = 1, yielding a,b,c = 5,1,94.

Answer 5

Your answer , 5, 1 and 94 do not satisfy the 2nd equation (10a + 3b + 5c = 100). You may need to check your question again.

Answer 6

a million/. Substituting the perfect equation into the different 2 supplies x + 4x + a million + 4z = 14 4x + a million - x + 2z = 8.5. both equations simplify, to 5x + 4z = 13 3x + 2z = 7.5. we now have were given it right down to a gadget of two equations. ideal to unravel this pair by technique of eliminating. Multiply the second one equation by by technique of two to get 5x + 4z = 13 6x + 4z = 15. Subtracting the first equation from the second one eliminates z, and grants x = 2. Substituting x = 2 into 5x + 4z = 13 supplies 5*2 + 4z = 13, or 10 + 4z = 13, and hence 4z = 3, and dividing by by technique of four supplies z = 0.seventy 5. And y = 4x + a million = 4*2 + a million = 8 + a million = 9. So the answer's (x,y,z) = (2,9,0.seventy 5). - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 2/. again we modify the perfect equation into the different 2, to get x + y + 4x = 2.5 x - y + 4x = 2.5. those equations simplify right down to 5x + y = 2.5 5x - y = 2.5 Subtracting the second one equation from the first supplies 2y = 0 and hence y = 0. Substituting y = 0 into 5x + y = 2.5 supplies 5x = 2.5, and dividing by technique of 5 supplies x = 0.5. And z = 4x = 4*0.5 = 2. So (x,y,z) = (0.5,0,2).

Answer 7

Are you sure you copied down the problem right? Looking at the graph for these two equations, we can see two planes that intersect at a line. Therefore, the solution is a line, not a point. The answer should be the equation of a line. Check to see if you copied down the answer right.

Answer 8

You have an underdetermined problem. There are an infinite number of solutions. You could, if you want, solve for the minimum norm solution (pseudoinverse(A)*B if the system of equations is rewritten in the form Ax = B). In Matlab, A\B gives a = 5.2632, b = 0, and c = 94.7368, and pinv(A)*B gives a = -6.3574, b = 44.1581, and c = 62.1993.

Answer 9

I have to agree with most of these people. Even if you isolate the first equation into a=100-b-c, b=100-a-c and c=100-a-b and sub them into the other equation you will be doing it forever. There has to be another element to this question.

Answer 10

You can't, but you can select a random number for a, b, or c and find the remaining two. In your answer you need to include: "Assumption: a=1" or something like that. Plug in your assumption to solve for b and c.