#### (New question of the week)

Now that we are receiving questions at the new site, we will be periodically posting some of those questions and answers, in addition to going back to particularly interesting questions from the old archive.

This question (which I have slightly edited) came to The Math Doctors earlier this month, from Navneet, who has asked many interesting questions in the past:

Consider this problem and its solution: "We have two strings A and B of lengths 5 cm and 6 cm respectively. And, we have another piece of string C whose length is 4 cm. Now, we have to make the strings A and B equal in length by cutting the string C and adding it to A and B such that string C is completely used. Find what length of strings should be added to A and B?" Solution: Let the length of string added to A be "x" and the length of string added to B be "y". Then, 5+x=6+y => x - y = 1 ......(i) And, x + y = 4 .....(ii) On adding (i) and (ii) we get, =>x = 5/2 = 2.5 and y = 1.5 Here, you can clearly see that we were asked to find the value of two unknown variables. And, we were able to find the values of both the unknown variables ONLY by using two different equations [(i) & (ii)]. So, my question is why is the number of unknown variables (whose value is to be found out) same as the number of equations required to find the value of those unknown variables? Is there any Mathematical reason behind this? I will be thankful for help!

Note that although he started with a particular application problem (which could be worth discussing in its own right), the focus of his question is on the nature of the system of equations used in solving it. As an aside, we have an old discussion of how the addition method of solving a system of equations relates (or not) to the physical quantities and units in the original problem: Subtracting Different Types of Units.

Here is my answer:

```
Hi, Navneet.
There are several ways to answer this question, or any "why" question.
In this case, one way is geometrical. A system of (linear) equations can be represented on a graph as a set of lines, planes, or higher-dimensional entities. Keeping ourselves to two variables, each equation represents a line, and the solution is the point or points where these lines intersect.
If there were only one equation, there would be infinitely many points that satisfy "all the equations" -- all the points on that one line.
If there were three equations, it would be very unlikely that they would all meet in one place; and you would only need two of them to find the point of intersection.
Two equations (two lines) are sufficient to find one point on a plane, because any two (non-parallel) lines intersect in one point.
In three dimensions, each equation represents a plane. Two planes typically intersect along a line. You need three planes to determine a single point.
Algebraically, you can just think about the method you use to solve. At each step, you combine two equations to eliminate a variable, so for example if you start with three equations, the first step eliminates one variable and one equation, the next step does the same, and you are now down to one equation in one variable, which you can solve. If you had a different number of variables, this would not have happened.
As I said, there are different ways to think about it any "why" question. If these two perspectives don't work for you, let us know what sort of answer are you hoping for.
Doctor Peterson
```

There is a lot more that could be said; Navneet was satisfied with this answer, so we didn’t continue the conversation. If he had wanted to dig deeper, I might have referred to these pages on Ask Dr. Math that deal with related issues:

Doctor Douglas, 2003: Number of Equations Needed in a Simultaneous Linear System

This is essentially the same as Navneet’s question, with a fuller answer.

Doctor Pete, 1997: Inconsistent and Dependent Systems Doctor Peterson, 2017: Inconsistent and Dependent?

These discuss the special cases in which the “right” number of equations don’t work; the second has links to more.

Doctor Rick, 2005: Sketching a Plane in Three Dimensional Space

This may help if you struggle with the three-dimensional case.

## A little extra

Let’s go back to Navneet’s application. Can we solve it without algebra? How might a non-algebraic method be related to the algebra?

"We have two strings A and B of lengths 5 cm and 6 cm respectively. And, we have another piece of string C whose length is 4 cm. Now, we have to make the strings A and B equal in length by cutting the string C and adding it to A and B such that string C is completely used. Find what length of strings should be added to A and B?"

Here is my immediate thought: We need to add 1 extra centimeter of string to A in order to make them equal; so imagine first taking 1 cm from C and adding it to A to make 6 cm. Then take the remaining 3 cm of C and divide it into two equal parts, each 1.5 cm. That means we’ve added 2.5 cm to A, and 1.5 cm to B.

In Navneet’s algebraic solution, he had the equation \(x – y = 1\), which amounts to my first observation. When he adds the equations, he is adding that 1 cm to the 4 cm of C; but if he had subtracted the equations instead (solving for \(y\) first, which amounts to what I did), he would have subtracted 1 from the 4, just as I did. (Can you make up a non-algebraic method that would be more like his algebraic solution?)

I find it interesting to look for non-algebraic solutions; in helping younger students who are not ready for algebra, I sometimes find that the algebraic method that comes more naturally to me can be translated into a more visual method that can work for kids. Algebra, as I often comment, is a way to save the effort of thinking through each problem in its own way, by turning all problems into equations that can be solved by routine. On the other hand, non-algebraic methods are more challenging, and therefore more fun!