Problem 31 - Calculus for Biology and Medicine

Question

Determine whether each system is overdetermined or underdetermined; then solve each system. $$ \begin{array}{r} 2 x-y=3 \\ x-y=4 \\ 3 x-y=1 \end{array} $$

Step-by-Step Solution

Answer

The system is overdetermined and inconsistent, yielding no solution.

Step 1: Understand the System

We are working with a system of linear equations:1. \(2x - y = 3\)2. \(x - y = 4\)3. \(3x - y = 1\)This system has three equations and two variables, namely \(x\) and \(y\). This indicates that the system is overdetermined because there are more equations than variables.

Step 2: Check for Consistency

To determine whether there is no solution, one solution, or infinitely many solutions, we need to check the consistency of the system. We'll use the method of substitution or elimination to check. Begin by solving the first two equations together.From equation (2): \(x - y = 4\) , we express \(x\) in terms of \(y\):\[ x = y + 4 \]

Step 3: Substitute and Simplify

Substitute \(x = y + 4\) into equation (1):\[ 2(y + 4) - y = 3 \]Simplify:\[ 2y + 8 - y = 3 \]\[ y + 8 = 3 \]\[ y = 3 - 8 \]\[ y = -5 \]

Step 4: Solve for x

Using \( y = -5 \), substitute back into \(x = y + 4\):\[ x = -5 + 4 \]\[ x = -1 \]Thus, from equations (1) and (2), we find \( x = -1 \) and \( y = -5 \).

Step 5: Verify Solutions with Third Equation

Substitute \(x = -1\) and \(y = -5\) into equation (3):\[ 3(-1) - (-5) = 1 \]\[ -3 + 5 = 1 \]\[ 2 eq 1 \]The solution \(x = -1, y = -5\) does not satisfy the third equation. This inconsistency indicates that the system is inconsistent.

Step 6: Conclusion

After verifying with all equations, we see that not all values hold for the given equations. Because the solution found from two equations does not satisfy the third equation, the system of equations is inconsistent, which means it has no solution.

Key Concepts

System of Linear Equations

A system of linear equations consists of multiple linear equations that have common variables. In the given exercise, we have a system with three equations:

• \(2x - y = 3\)
• \(x - y = 4\)
• \(3x - y = 1\)

Here, the variables are \(x\) and \(y\). Systems can generally be classified based on the number of equations and the number of variables.
In this case, we have more equations (3) than variables (2), which leads us to categorize it as an overdetermined system. This means that there could be conflicting information in the equations, leading to potential inconsistencies. Such systems may not have a solution that satisfies all equations.

Consistency of Equations

Consistency in a system of equations is about whether or not there is at least one set of values for the variables that satisfies all the equations simultaneously.
A system can be consistent, having one or several solutions, or inconsistent, having no solutions at all.
To test for consistency, methods like substitution can be employed to deduce potential solutions.
In our exercise, solving the first two equations gives us values \(x = -1\) and \(y = -5\).
However, substituting these values into the third equation shows a discrepancy:

• The calculated sum is \(2\), not the required \(1\).

This indicates that all three equations cannot be simultaneously satisfied, marking the system as inconsistent and suggesting it has no viable solutions.

Solution of Linear Systems

Solving a linear system aims at finding values for the variables that make all the equations true simultaneously.
In simple cases, techniques like substitution or elimination are used to reduce equations, finding explicit solutions.
For this system, attempting to solve for variables \(x\) and \(y\) using different equations may yield a solution, like \(x = -1\) and \(y = -5\), which solves some but not all equations.
Ultimately, when checking against all provided equations, these solutions must satisfy each one to conclude consistency.
When this does not occur, as shown by failing to validate against the third equation, we conclude the system is unsolvable as an overdetermined system.
This inherently means there are no solutions since the set of equations in the system are in conflict.