How Do You Solve An Application That Results In A Dependent System Of Equations?

Some of the most interesting problems to solve are problems that lead to a system of linear equations where there are more variables than equations. Typically, these systems have many possible equations. Figuring out which of those solutions make sense and which do not is challenging.

Here is one such example

A restaurant owner orders a replacement set of knives, fork, and spoons. The box arrives containing 40 utensils and weighing 141.3 ounces (ignoring the weight of the box). A knife, fork, and spoon weigh 3.9 ounces, 3.6 ounces, and 3.0 ounces, respectively.

How many knives, forks and spoons are in the box?

Since we are being asked to find the number of knives, forks, and spoons, let’s make the following designations:

K: number of knives

F: number of forks

S: number of spoons

The first thing to notice is that you are given a total number of utensils (40) and a total weight for the utensils (141.3 ounces). These are the prime candidates for writing out the equations.

Starting with

Total number of utensils = 40

It is fairly obvious that

K + F + S = 40

Starting with

Total weight of utensils = 141.3

We can deduce that the individual weights are

Weight of knives = 3.9 K

Weight of forks = 3.6 F

Weight of spoons = 3.0 S

So

3.9 K + 3.6 F + 3.0 S = 141.3

Now what? Your experience in these sections probably tells you that you need another equation in the three unknowns to be able to solve for K, F, and S. This is true if there is a unique solution to this problem. But this problem actually has many possible solutions (ie. There are many ways to have 40 utensils that weigh 141.3 ounces).

So let’s simply solve the problem as is with Gauss-Jordan elimination:

K + F + S = 40

3.9 K + 3.6 F + 3.0 S = 141.3

Start by converting this to an augmented matrix:

To put into row echelon form, let’s multiply the first row by -3.9 and add it to the second row. Put the result in row 2:

To make the first nonzero number in the second row a 1, multiply the entire row by 1/-0.3 and replace the second row with that result:

This is now in row echelon form. It is almost as if there was a third row in the matrix, but it is all zeros.

With one more row operation, we can put the system in reduced row echelon form.

Converting this back to a system of equations, we get

K – 2S = -9

F + 3S = 49

To get our solutions, the first equation for K and the second equation for F:

K = 2S – 9

F = -3S+49

This is pretty easy to do since the system was in reduced row echelon form.

The variable S can be any value that makes sense for the problem. For instance, we certainly know that S (the number of spoons) should be non-negative integers like S = 0, 1, 2, … If we make up a table, we can make some interesting observations:

We need the numbers of each type of utensil to be positive…to do that we’ll require the number of spoons to be S = 5, 6, 7, …, 16. So even though the system has an infinite number of solutions, only certain ones make sense. In other words, any ordered triple of the form (K, F, S) = (2S – 9, -3S + 49, S) as long as S is an integer from 5 to 16.

How Do I Solve A System Of Equations That Results In Two Equations And Three Variables?

When using Left to Right Elimination or Gauss’s Methods, a system of three equation in three variables often results in a unique solution. In other words, a single value for each of the three variables solves each of the three equations in the system. However when some systems are put into row echelon form, all of the constants and variables in one of the equations drop out. In a sense, one of the equations has disappeared and now you are left with a system of two equations in three variables.

Don’t confuse this with an inconsistent system. In an inconsistent system, the row operations lead to one of the equations being false…something like 0 = 6. This type of system has no solutions.

Let’s examine a system of three equations in three variables in which one of the equations drops out. Take the system

\displaystyle \begin{matrix}  x&+2y&-4z &=5 \\  2x&+3y&-z &=3 \\  3x&+5y&-5z &=8 \\  \end{matrix}

If we use left to right elimination to solve this system, we get the equivalent system

\displaystyle \begin{matrix}  &x&+2y&-4z &=5 \\  &&y&-7z &=7 \\  & & &0 &=0 \\  \end{matrix}

This new system contains only two equations since 0 = 0 does not describe a relationship between the variables. This new system has many solutions. We can show this by solving the second equation for y to give

\displaystyle y=7z+7

and substituting it into the first equation:

\displaystyle x+2\left( 7z+7 \right)-4z=5

Now solve this equation for x,

\displaystyle \begin{matrix}  x+14z+14-4z=5 \\  x+10z+14=5 \\  x=-10z-9 \\  \end{matrix}

The solution to the system is

\displaystyle \begin{matrix}  x&=&-10z-9 \\  y&=&7z+7 \\  \end{matrix}

This may also be written as an ordered triple \left( -10z-9,7z+7,z \right).  We have solved for x and y, but z is not specified. We can pick it to be anything. For instance, if  z = 0, then x = -0 and y = 7. This means that the ordered triple (-9, 7, 0) solves the original system of equations. If z = -1, then x = 1 and y = 0, So (1, 0, -1) also solves the system.