Problem 23 - Calculus for Biology and Medicine

Question

In Problems 19-24, solve each system of linear equations. $$ \begin{array}{l} 2 x-y+3 z=3 \\ 2 x+y+4 z=4 \\ 2 x-3 y+2 z=2 \end{array} $$

Step-by-Step Solution

Answer

The solution is dependent: \(x = \frac{7}{2}t\), \(y = t\), \(z = 1 - 2t\) for any real number \(t\).

Step 1: Organize the System of Equations

The system of linear equations is given by: \(2x - y + 3z = 3\), \(2x + y + 4z = 4\), and \(2x - 3y + 2z = 2\). Ensure coherence and identify they all have the same \(x\) term that can help in elimination.

Step 2: Eliminate x by Subtracting Equations

Subtract the first equation from the second: \((2x + y + 4z) - (2x - y + 3z) = 4 - 3\). This simplifies to \(2y + z = 1\). Name this Equation (4).

Step 3: Eliminate x from Another Pair

Subtract the first equation from the third: \((2x - 3y + 2z) - (2x - y + 3z) = 2 - 3\). This simplifies to \(-2y - z = -1\). Name this Equation (5).

Step 4: Solve the Two-Variable System

We have two equations: (4) \(2y + z = 1\) and (5) \(-2y - z = -1\). Add these two equations to eliminate \(z\): \((2y + z) + (-2y - z) = 1 - 1\), leading to \(0 = 0\), which suggests the system may have multiple solutions or be dependent.

Step 5: Check for Consistency and Substitute

Solve for one variable in terms of another based on the simpler Equation (4): \(z = 1 - 2y\). Substitute into one original equation, like the first: \(2x - y + 3(1 - 2y) = 3\) leading to \(2x - 7y + 3 = 3\) simplifying to \(2x = 7y\).

Step 6: Parameterize the Solution

Since the solution yielded dependencies between variables: let \(y = t\), then \(x = \frac{7}{2}t\), and \(z = 1 - 2t\), where \(t\) is any real number. Thus, the solutions are parameterized as functions of \(t\).

Key Concepts

System of Equations

A system of equations involves solving for multiple unknowns. In our case, we have three unknowns: \(x\), \(y\), and \(z\). The system of linear equations is expressed in the form of:

• \(2x - y + 3z = 3\)
• \(2x + y + 4z = 4\)
• \(2x - 3y + 2z = 2\)

These equations must be solved simultaneously to find values of \(x\), \(y\), and \(z\) that satisfy all three equations at once. It's crucial to organize and manipulate the equations to reveal hidden relationships among unknowns. When solving, we often aim to eliminate one variable to simplify our work.

Solution Methods

There are various solution methods to tackle a system of linear equations like substitution, elimination, or using matrices. For our system:

• We used the elimination method to remove \(x\) initially.
• This allowed us to handle a simpler 2-variable system (in our case, equations with only \(y\) and \(z\)).

The elimination method usually involves subtracting one equation from another to eliminate a variable. This technique can make a complex problem more manageable by reducing the number of unknowns. Remember, the choice of method depends on the structure of the equations and often on personal preference.

Parameterization

Parameterization refers to expressing the solutions of a system of equations in terms of one or more independent variables (parameters). As we've seen, by introducing \(t\) as a parameter, we can express:

• \(y = t\)
• \(x = \frac{7}{2}t\)
• \(z = 1 - 2t\)

Parameterization is particularly useful in infinite solution cases or dependent systems because it provides a way to describe all possible solutions compactly. By setting one variable freely, we can derive relationships between the others.

Dependent Systems

Dependent systems are those where the equations in the system are not independent from each other. In other words, one or more equations are linear combinations of others. In our exercise:

• The solutions led us to the identity \(0 = 0\) during the elimination process.
• This indicates that the system of equations has infinite solutions.

This happens when equations describe planes that coincide or are parallel. Dependent systems require careful interpretation and use of parameters to express the myriad of solutions. It's essential to recognize that not all systems will have neat, singular solutions; some reflect more complex interconnections.