Section 14.1 Systems of Equations in Action
Imagine you are in a house with five roommates and are trying to divide up collective chores. Among yourselves it has been agreed that there are 20 total hours of chores per week. Additionally, there are some outside deals and requirements made between roommates. Let \(x_1, x_2, x_3, x_4, x_5\) denote the number of chore hours allotted to each roommate.
Subsection 14.1.1 Initial Constraints
Initially, the roommates propose the following conditions for chore distribution:
-
The total chore hours must be 20:\begin{equation*} x_1 + x_2 + x_3 + x_4 + x_5 = 20 \end{equation*}
-
Roommates 1 and 5 are a couple and want to maximize time together, so they insist on having the same number of chore hours:\begin{equation*} x_1 = x_5 \Rightarrow x_1 - x_5 = 0 \end{equation*}
-
Roommate 5 agrees to take on \(3\) more hours than roommate 3, in exchange for access to roommate 3’s car:\begin{equation*} x_5 = 3 + x_3 \Rightarrow x_5 - x_3 = 3 \end{equation*}
-
Roommate 3 wants to only work \(2\) hours in exchange for doing the more physically intense tasks:\begin{equation*} x_3 = 2 \end{equation*}
-
Roommate 5 has a fixed work schedule and can only contribute 4 hours per week:\begin{equation*} x_5 = 4 \end{equation*}
We can combine these conditions and form the following system of equations:
\begin{equation*}
\left\{\begin{array}{l}
x_1 + x_2 + x_3 + x_4 + x_5 = 20 \\
x_1 - x_5 = 0 \\
- x_3 + x_5 = 3 \\
x_3 = 2 \\
x_5 = 4
\end{array}\right.
\end{equation*}
We can then find the distribution hours each roommate should be given to honor every deal by using the
solve function:
We can see that the resulting system is inconsistent because the
solve function returned an empty list. Therefore a new set of deals must be made as these deals can not all be honored at the same time.
Subsection 14.1.2 Renegotiated System
After further discussion everyone has agreed to the following new set of deals:
-
The total chore hours still must be 20:\begin{equation*} x_1 + x_2 + x_3 + x_4 + x_5 = 20 \end{equation*}
-
Roommates 1 and 5 still want equal chore hours:\begin{equation*} x_1 - x_5 = 0 \end{equation*}
-
Roommate 5 is still assigned \(3\) more hours than roommate 3:\begin{equation*} x_5 - x_3 = 3 \end{equation*}
-
Roommate 2 pays a lower portion of rent and the group has agreed that their chores should be \(4/3\) of the average of the other roommates:\begin{equation*} \frac{4}{3} \cdot \frac{x_1 + x_3 + x_4 + x_5}{4} = x_2 \Rightarrow x_1 - 3x_2 + x_3 + x_4 + x_5 = 0 \end{equation*}
This new set of conditions forms the following system of equations:
\begin{equation*}
\left\{\begin{array}{l}
x_1 + x_2 + x_3 + x_4 + x_5 = 20 \\
x_1 - x_5 = 0 \\
- x_3 + x_5 = 3 \\
x_1 - 3x_2 + x_3 + x_4 + x_5 = 0
\end{array}\right.
\end{equation*}
We can then check for the existence of solutions again by using
solve again:
This system of equations has infinitely many solutions as seen by the parameterization \(x_5 = r_1\text{:}\)
\begin{equation*}
\left\{\begin{array}{l}
x_1 = r_1 \\
x_2 = 5 \\
x_3 = r_1 - 3 \\
x_4 = - 3r_1 + 18 \\
x_5 = r_1
\end{array}\right.
\end{equation*}
We must have \(3 \le r_1 \le 6\) because other values would result in negative chore hours due to rows 3 and 4.
Subsection 14.1.3 A Unique Solution
Roommate 5 brings up their fixed schedule again and the group agrees to include it. So we must add the following to the system of equations:
\begin{equation*}
x_5 = 4
\end{equation*}
This results in the following matrix and system:
\begin{equation*}
\left\{\begin{array}{l}
x_1 + x_2 + x_3 + x_4 + x_5 = 20 \\
x_1 - x_5 = 0 \\
- x_3 + x_5 = 3 \\
x_1 - 3x_2 + x_3 + x_4 + x_5 = 0 \\
x_5 = 4
\end{array}\right.
\end{equation*}
We can check for constistancy by using
solve again in sage:
We now obtain the unique solution:
\begin{equation*}
x_1 = 4,\ x_2 = 5,\ x_3 = 1,\ x_4 = 6,\ x_5 = 4
\end{equation*}
