Problem 46 - Intermediate Algebra

Question

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 2 x+2 y+3 z=10 \\ 3 x+y-z=0 \\ x+y+2 z=6 \end{array}\right. $$

Step-by-Step Solution

Answer

The system is dependent with infinite solutions.

Step 1: Identify the System of Equations

We have a system of three linear equations with three variables: \(x\), \(y\), and \(z\). The equations are:1. \(2x + 2y + 3z = 10\)2. \(3x + y - z = 0\)3. \(x + y + 2z = 6\).

Step 2: Eliminate \(y\) from Equations

First, we want to eliminate \(y\) from two pairs of equations. We'll use equations 1 and 2:1. From equation 1, simplify to \(y = 5 - x - 1.5z\).2. Substitute \(y\) in equation 2: \(3x + (5 - x - 1.5z) - z = 0\). This simplifies to \(2x - 2.5z + 5 = 0\).

Step 3: Solve for \(x\) and \(z\)

From the simplified equation \(2x - 2.5z + 5 = 0\), express \(x\) in terms of \(z\): \(x = 1.25z - 2.5\).Substitute this expression in equation 3 to eliminate \(x\):\((1.25z - 2.5) + y + 2z = 6\) leads to \(y = 8.75 - 3.25z\).

Step 4: Substitute Back to Find \(z\)

Substitute \(x = 1.25z - 2.5\) and \(y = 8.75 - 3.25z\) into the modified equation 3:\(1.25z - 2.5 + 8.75 - 3.25z + 2z = 6\).Simplify: \(0 = 0\).This shows the equations are consistent and dependent.

Step 5: Interpret the Dependency

Since the equation \(0 = 0\) holds true, the system has an infinite number of solutions, indicating that the equations are dependent. Thus, any \(z\) value substituted into the expressions \(x = 1.25z - 2.5\) and \(y = 8.75 - 3.25z\) will satisfy the system.

Key Concepts

Dependent Systems

A dependent system of equations means that each equation provides the same data in different forms. It's like describing the same object using different perspectives. Therefore, all equations essentially represent the same line or plane in a graphical view.

When we say equations are dependent, we mean:

• They rely on each other, and one is a linear combination of the others.
• They don’t offer new information. Instead, they confirm the data already described by the other equations.
• This results in the system being consistent, as there is no contradiction between equations.

In the context of a 3D space, dependent systems often represent the same plane. Hence, any point on this plane satisfies all the equations. If you find two equations multiplying various numbers, but simplifying to the same relationship, they are dependent.

Linear Equations in Three Variables

Linear equations in three variables are solved to find the values of three unknowns. They are represented as a system of equations and usually involve variables like \( x \), \( y \), and \( z \).

Each equation is a flat surface in three-dimensional space called a plane:

• The equation \( 2x + 2y + 3z = 10 \) is one such plane.
• Similarly, \( 3x + y - z = 0 \) and \( x + y + 2z = 6 \) represent other planes.

The core idea is to manipulate these equations to eliminate variables and understand how these planes intersect:

• If the planes intersect in a line or coincide completely, the system is dependent.
• If they intersect at a single point, this gives a unique solution.

The key technique often involves using substitution and elimination methods to simplify and solve the system effectively.

Infinite Solutions to Linear Equations

Infinite solutions occur when equations in the system aren't independent. Instead, they overlap, providing numerous solutions that satisfy all the equations.

When solving such a system, you might find:

• An expression simplifies to an identity, such as \( 0 = 0 \), after variable elimination.
• This indicates a true statement with no unique variable constraint, confirming infinite solutions.

Let's break down how we reach these solutions:You work through the system by eliminating variables:

• When you simplify the equations and find no specific solution for a variable, substitute different values for one variable to find corresponding values for others.

For the given system, equations simplify to valid identities that demonstrate infinte solutions. Thus, you can choose any suitable value for \( z \) and calculate corresponding \( x \) and \( y \), providing endless solution sets.