Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the system of equations $$\begin{eqnarray} 3a &=& A \\ 4b &=& B \\ 7c&=& C \\ a+b+c&=&S \end{eqnarray}$$ subject to $$A>S,\quad B>S,\quad C>S$$

How can I find $a$, $b$ and $c$?

share|improve this question
Are the variables integers, or may they be reals? There are divisibility tests available if integers. –  Ross Millikan May 1 '12 at 13:25
They are from Q –  Rma May 2 '12 at 15:18
add comment

2 Answers

up vote 3 down vote accepted

I assume that $S$ is given and fixed, while we search for $a,b$ and $c$. To find a solution just draw a picture:

Start with the three axes of a coordiante system. The triangle with the corners $(0,0,S)$, $(0,S,0)$ and $(S,0,0)$ represents all solutions of the equation $a+b+c=S$.

Now you want to enforce the inequalities, i.e. $a>S/3$, $b>S/4$ and $c>S/7$ (we got rid of $A,B,C$ here). Each of these conditions can be represented by a plane parellel to a coordinate plane with the given distance. For example for $a>S/3$ you draw a plane parallel to the $b$-$c$-plane, with distance $S/3$. You will cut out a part of the original triangle and thereby find all solutions. It is an easy exercise to explicitly compute the intersections of the planes and the triangle.

share|improve this answer
and just for curiosity, if a is given, how can I find out b and c ? Thank you for your answer –  Rma May 1 '12 at 19:18
add comment

Writing $v=(a,b,c)$, we have (by subtracting out LHS and then collecting terms)

$$\begin{cases}3a>a+b+c \\ 4b>a+b+c \\ 7c>a+b+c \end{cases} \implies \begin{cases}v\cdot (2,-1,-1)>0 \\ v\cdot (-1,3,-1)>0 \\ v\cdot (-1,-1,6)>0\end{cases}$$

Each of the dot product equations defines a halfspace in $\Bbb R^3$; combine them and they create what I guess we can think of as an "infinite pyramid" or "pyramid without base" or "pyramidal cone." In any case, it has three edges which are rays outwards from the origin. These are computed as the pairwise intersections of the three given planes. Intersections can be computed by taking the cross product of the normal vectors of the two planes; the intersection is perpendicular to both after all.

(This is just some geometric intuition behind the solution set.)

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.