Problem 50 - College Algebra

Question

For the following exercises, solve the system for \(x, y,\) and \(z\) $$ \begin{aligned} \frac{x-1}{3}+\frac{y+3}{4}+\frac{z+2}{6} &=1 \\ 4 x+3 y-2 z &=11 \\ 0.02 x+0.015 y-0.01 z &=0.065 \end{aligned} $$

Step-by-Step Solution

Answer

The system has no solution for \( x, y, z \) since it's inconsistent.

Step 1: Simplify the First Equation

The first equation is \( \frac{x-1}{3} + \frac{y+3}{4} + \frac{z+2}{6} = 1 \). Clear the fractions by multiplying the entire equation by 12. This gives \( 4(x-1) + 3(y+3) + 2(z+2) = 12 \). Simplifying, we get \( 4x + 3y + 2z = 23 \).

Step 2: Set Up the System of Equations

Now, write down the system of equations: \[ \begin{aligned} 4x + 3y + 2z &= 23 \ 4x + 3y - 2z &= 11 \ 0.02x + 0.015y - 0.01z &= 0.065 \end{aligned}\]

Step 3: Eliminate a Variable

Subtract the second equation from the first to eliminate \( z \). This gives: \[ (4x + 3y + 2z) - (4x + 3y - 2z) = 23 - 11 \] Simplifying, we get \( 4z = 12 \), so \( z = 3 \).

Step 4: Substitute for \( z \)

Substitute \( z = 3 \) into the second equation, \( 4x + 3y - 2z = 11 \). This becomes \( 4x + 3y - 6 = 11 \). Simplify to \( 4x + 3y = 17 \).

Step 5: Solve for Another Variable

Substitute \( z = 3 \) into the third equation, \( 0.02x + 0.015y - 0.01z = 0.065 \). This becomes \( 0.02x + 0.015y - 0.03 = 0.065 \). Simplify to \( 0.02x + 0.015y = 0.095 \). Divide by 0.005 to simplify: \[ 4x + 3y = 19 \].

Step 6: Solve the Resulting System

Now solve the system:\[ \begin{aligned} 4x + 3y &= 17 \ 4x + 3y &= 19 \end{aligned}\]The equations are inconsistent since both can't be true (i.e., one system suggests one thing while the other suggests another), so the system has no solution for \( x \) and \( y \).

Key Concepts

Linear Equations

Linear equations are mathematical statements of equality involving one or more variables. They take the form of a straight line when graphed in a coordinate system. Each term in a linear equation can either be a constant or a multiple of the variable. Common examples include expressions like \( ax + by + cz = d \), where \( a, b, \) and \( c \) are coefficients, \( x, y, \) and \( z \) are variables, and \( d \) is the constant term.
Linear equations are foundational in algebra and are used to describe relationships between variables. They're called 'linear' because the graph of the solution set forms a line. In the context of systems of equations, we often deal with multiple linear equations simultaneously, trying to find values for the variables that satisfy all equations in the system.

• Each equation represents a line or plane in multidimensional space.
• The solution is the point or set of points where the lines or planes intersect.

Understanding linear equations is crucial for solving more complex algebraic problems, as they provide a straightforward way to model relationships and constraints.

Variable Elimination

Variable elimination is a strategic approach used to simplify and solve systems of equations. The goal here is to remove one or more variables by combining the equations in such a way that only one variable remains, which can then be easily solved for. This technique is especially employed in systems of linear equations.
Let's consider our example system of equations:

• The first step is to choose which variable to eliminate.
• We can add or subtract the equations to remove a chosen variable, allowing us to simplify the system step by step.

In our exercise, we eliminated \( z \) through subtraction. Specifically:

• The expression \( (4x + 3y + 2z) - (4x + 3y - 2z) = 23 - 11 \) allowed us to find \( z = 3 \).

This technique is powerful because it reduces the complexity of the system, making it easier to find solutions for the remaining variables.

Inconsistent Systems

An inconsistent system is a type of system of equations in which no set of values for the variables will satisfy all the equations simultaneously. This occurs when the lines or planes formed by the equations do not intersect at any common point. They may be parallel or divergent.
Within the solved exercise, we encountered an inconsistent system while attempting to find solutions for \( x \) and \( y \). We ended up with equations:

• \( 4x + 3y = 17 \)
• \( 4x + 3y = 19 \)

These two equations suggest conflicting values for the same expression, indicating there is no possible solution that satisfies both simultaneously.
This highlights a critical aspect of solving systems of equations: not all systems have solutions. Understanding this nature of inconsistency helps to prevent endless calculations and ensures any system-solving strategies are adapted accordingly.

Substitution Method

The substitution method is a technique for solving systems of equations that involves solving one of the equations for one variable and then substituting that expression into the other equations. This can be especially useful when dealing with simpler systems or when direct elimination is not feasible.
When you use substitution, you effectively reduce the number of variables in an equation, simplifying the process of finding a solution. The steps generally involve:

• Solving one equation for one variable in terms of others.
• Substituting this expression into the other equations.

In our exercise, after finding \( z = 3 \), substitution was used to simplify the other equations. For example, we substituted \( z \) into the second equation, transforming it to \( 4x + 3y = 17 \). This substitution progressively reduced the complexity of the equation set.
Though ultimately leading to an inconsistency, the substitution method is still a powerful approach in systems where elimination might be less clear-cut or where equations are already partially solved.