Problem 34 - Precalculus: Mathematical for Calculus

Question

Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, find the complete solution. $$\left\\{\begin{aligned} y-5 z &=7 \\ 3 x+2 y &=12 \\ 3 x &+10 z=80 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The system is inconsistent.

Step 1: Write the system of equations

We start with the given system of equations: 1. \( y - 5z = 7 \) 2. \( 3x + 2y = 12 \) 3. \( 3x + 10z = 80 \) These equations will be used to determine if there is a solution, and if so, what type it is.

Step 2: Eliminate variables method

To understand the nature of this system, let's eliminate one of the variables and simplify.To eliminate \( x \), set the equations 2 and 3 equal since they both have \( 3x \) terms:- From equation 2, solve for \( 3x = 12 - 2y \).- From equation 3, solve for \( 3x = 80 - 10z \).Equating both expressions gives: \[ 12 - 2y = 80 - 10z \] Simplifying, we get: \[ 2y + 10z = 68 \] Divide by 2 to further simplify: \[ y + 5z = 34 \]

Step 3: Compare simplified equations

We now compare the simplified equation from Step 2, \( y + 5z = 34 \), with the original equation 1, \( y - 5z = 7 \). They are different, and thus the system is inconsistent.Adding them gives: \( (y + 5z) + (y - 5z) = 34 + 7 \) \( 2y = 41 \) This results in a contradiction when attempting to find a consistent pair for each equation in the system.

Key Concepts

Inconsistent System

An inconsistent system is a set of equations with no solutions. In other words, there is no single combination of variables that satisfies all equations simultaneously. In the context of linear equations, inconsistency arises when the lines represented by the equations do not intersect at any point.

Consider the given system of equations:

• Equation 1: \( y - 5z = 7 \)
• Equation 2: \( 3x + 2y = 12 \)
• Equation 3: \( 3x + 10z = 80 \)

To determine consistency, we simplify the system by eliminating variables. This leads to comparing simplified equations. In this particular exercise, the simplification results in equations \( y + 5z = 34 \) and \( y - 5z = 7 \).

Notice how both equations have different y-intercepts which mean they will never intersect. Thus, the system is inconsistent.

Dependent System

A dependent system, unlike an inconsistent one, has infinite solutions. This occurs when all equations describe the same line, and therefore, intersect at every point along the line.

For this scenario, the system would reduce to identical equations after simplification. In essence, each equation can be derived from the others by multiplication, addition, or subtraction by a constant.

However, in the given exercise: after elimination and simplification, the new equation \( y + 5z = 34 \) is not identical to the other existing equation \( y - 5z = 7 \). Rather than resulting in identical lines, these are distinct, suggesting the system is not dependent but inconsistent.

Remember, whenever a set of linear equations is dependent, it typically reduces down to a single equation or multiple ones dependent on each other.

Eliminating Variables

The method of eliminating variables is a strategic approach used to simplify a system of equations. By focusing on removing one variable at a time, you can reduce the complexity and make solving for the remaining variables easier.

In our example, to eliminate \( x \), we set equations 2 and 3 equal because both contain \( 3x \):

• From equation 2, solve for \( 3x = 12 - 2y \)
• From equation 3, solve for \( 3x = 80 - 10z \)

Setting these expressions equal to one another results in \( 12 - 2y = 80 - 10z \), leading to \( 2y + 10z = 68 \). When further simplified, we arrive at \( y + 5z = 34 \).

This new equation is then compared with the original equations to determine the nature of the system. The comparison showed differing solutions, indicating an inconsistency. Eliminating variables is crucial for clarifying potential solutions or lack thereof.