Problem 44 - Precalculus Mathematics for Calculus

Question

Solve the system of linear equations. $$\left\\{\begin{aligned} 3 x+y &=2 \\ -4 x+3 y+z &=4 \\ 2 x+5 y+z &=0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system of equations has no solution (inconsistent system).

Step 1: Simplify the System

The given system of equations is already simple enough, so we start with labeling the equations for reference:1) \(3x + y = 2\)2) \(-4x + 3y + z = 4\)3) \(2x + 5y + z = 0\).

Step 2: Eliminate z from Equations 2 and 3

Subtract equation 3 from equation 2 to eliminate \(z\):\((-4x + 3y + z) - (2x + 5y + z) = 4 - 0\).This leads to:\(-6x - 2y = 4\).

Step 3: Solve for y in Equation 1

Solve Equation 1 for \(y\):\(y = 2 - 3x\).

Step 4: Substitute y into New Equation

Substitute \(y = 2 - 3x\) into the new equation from Step 2:\(-6x - 2(2 - 3x) = 4\).This simplifies to:\(-6x - 4 + 6x = 4\).Simplify further to get \(-4 = 4\), which is a contradiction.

Step 5: Conclude No Solutions

Since we reached a contradiction, it means the system of equations does not have a solution as the lines represented do not intersect at any point.

Key Concepts

Solving Systems of Equations

Solving systems of equations is a fundamental aspect of algebra. It involves finding values for variables that satisfy all equations in the system. In our specific example, we are looking at a system of three linear equations. These equations form a system that could represent three lines in a three-dimensional space.

Here are common methods used to solve such systems:

• Graphically: By drawing each equation and identifying intersections.
• Substitution: Solving one equation for a variable and substituting it into the other equations.
• Elimination: Adding or subtracting equations to eliminate variables systematically.

The goal is to determine if there is a single solution (a point where all equations intersect), no solution (no common intersection point), or infinitely many solutions (all equations intersect along a line or plane). Step-by-step methods can be employed to examine these possibilities thoroughly.

Elimination Method

The elimination method is especially useful for systems where substitution might be cumbersome. In this approach, we carefully add or subtract entire equations to systematically remove one variable at a time. This is what we applied to our initial system of equations.

In our exercise:

• We labeled the equations for clarity and began by eliminating the variable \(z\) from two of the equations.
• This was done by subtracting one equation from another.
• This operation gave us a new equation with only the variables \(x\) and \(y\).

The beauty of the elimination method is that it reduces complexity step by step, making it easier to spot relationships among the remaining variables. It's like peeling away layers until the answer becomes readily apparent.

Contradiction in Systems of Equations

Sometimes, when solving a system using any one method, we may encounter a contradiction. A contradiction arises when an operation leads to a statement that is false and thus impossible. In our example, the step \(-4 = 4\) indicated a contradiction, meaning no possible values for \(x\) and \(y\) exist to satisfy the equations simultaneously.

A contradiction usually implies:

• The lines or planes represented by the original equations never meet; hence, there is no solution.
• The graphical interpretation shows the lines are parallel or planes are skewed, underscoring this lack of intersection.

Recognizing a contradiction is crucial. It saves time by conclusively revealing that pursuing further calculations won’t be fruitful and highlights the underlying nature of the system's relationships.