Problem 48 - Precalculus: Mathematics 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 because there are no solutions that satisfy all the equations simultaneously.

Step 1: Identify the system of equations

We have the following system of linear equations: \(y - 5z = 7\), \(3x + 2y = 12\), and \(3x + 10z = 80\). We need to determine whether the system is inconsistent, consistent and dependent, or consistent and independent.

Step 2: Solve the equations using substitution or elimination

First, we will solve the third equation for \(3x\) as \(3x = 80 - 10z\). Substitute \(3x = 80 - 10z\) into the second equation: \(80 - 10z + 2y = 12\). Simplify to obtain \(2y = 12 - 80 + 10z\), which results in \(2y = 10z - 68\). Dividing every term by 2, we have \(y = 5z - 34\).

Step 3: Check consistency between equations

Use the expression found for \(y\) in Step 2 in the first equation: \(y - 5z = 7\). Substitute \(y = 5z - 34\), resulting in \(5z - 34 - 5z = 7\). Simplifying, we get \(-34 = 7\), which is a contradiction, indicating there are no solutions that satisfy all equations simultaneously.

Step 4: Conclusion about the system

Since the contradiction in Step 3 shows that there is no solution that satisfies all equations, the system of equations is inconsistent.

Key Concepts

Linear Equations

Linear equations are a fundamental concept in algebra describing a straight line when graphed. They involve variables raised only to the power of one, producing an equation of the form \(ax + by = c\). Here, \(a\), \(b\), and \(c\) are constants, while \(x\) and \(y\) are variables. Linear equations can represent real-world situations where variables show a direct relationship. For example, in the given exercise, equations like \(y - 5z = 7\), \(3x + 2y = 12\), and \(3x + 10z = 80\) involve variables \(x\), \(y\), and \(z\). Each equation depicts a linear relationship between these variables. Understanding linear equations is crucial as they form the basis for solving problems involving systems of equations.

Substitution Method

The substitution method is a technique used to find the values of variables that satisfy a system of equations. This method involves solving one of the equations for one variable and then substituting that expression into another equation. The ultimate goal is to find a solution where all equations are satisfied simultaneously.
In our exercise, we used the substitution method starting with the equation \(3x + 10z = 80\). We expressed \(3x\) in terms of \(z\), as \(3x = 80 - 10z\), and then substituted into the equation \(3x + 2y = 12\). This allows us to write \(2y\) in terms of \(z\) and find a relationship between \(y\) and \(z\). By simplifying, we reached \(y = 5z - 34\). This method simplifies complex systems by reducing them to manageable steps.

Elimination Method

The elimination method is another common approach to solving systems of equations. Unlike substitution, this method involves adding or subtracting equations to eliminate one of the variables, making it possible to solve for the others. This is especially useful when dealing with equations where coefficients allow straightforward elimination of variables.
To apply the elimination method in our case, we would normally align equations to cancel out terms by altering coefficients. However, in the exercise, most steps were achieved using substitution due to the provided equation alignments. When solving systems with more variables, elimination serves as a versatile tool, often used interchangeably with substitution based on the complexity and structure of the equations involved.

Solution of Equations

Finding the solution to systems of equations involves determining values for each variable that satisfy all given equations simultaneously. Solutions can be classified as consistent or inconsistent. A consistent system has at least one solution, while an inconsistent system has none. A further division is made into dependent and independent solutions, where dependent systems have infinitely many solutions, while independent systems generally have a unique solution.
In this exercise, upon substituting the derived value of \(y = 5z - 34\) back into the first equation, \(y - 5z = 7\), we obtained a contradiction: \(-34 = 7\). This indicates the system is inconsistent as it cannot have a solution. Such systems often occur when the equations represent parallel lines with no point of intersection. Understanding such classifications is essential in algebra for analyzing real-world scenarios and solving mathematical problems accurately.