Problem 52 - Intermediate Algebra

Question

Solve each system, if possible. If a system is inconsistent or if the equations are dependent, state this. $$ \left\\{\begin{array}{l} 4 a-3 b=1 \\ 6 a-8 c=1 \\ 2 b-4 c=0 \end{array}\right. $$

Step-by-Step Solution

Answer

The solution is \(a = -0.5\), \(b = -1\), \(c = -0.5\).

Step 1: Analyze the Equations

We start by examining each of the three equations given:1. \(4a - 3b = 1\)2. \(6a - 8c = 1\)3. \(2b - 4c = 0\).We notice there are three variables (\(a, b,\) and \(c\)) and three equations. Our goal is to see if we can find values for these variables that satisfy all equations simultaneously.

Step 2: Solve the Third Equation for b

From the third equation, \(2b - 4c = 0\), we solve for \(b\):\[ 2b = 4c \]\[ b = 2c \]We express \(b\) in terms of \(c\).

Step 3: Substitute b in the First Equation

Substitute \(b = 2c\) into the first equation:\[ 4a - 3(2c) = 1 \]\[ 4a - 6c = 1 \]Now, we have \(4a - 6c = 1\).

Step 4: Make a New System with Two Equations

Now we have a reduced system of two equations:1. \(4a - 6c = 1\)2. \(6a - 8c = 1\).

Step 5: Solve the Reduced System

Solve the reduced system by elimination or substitution. First, multiply the first equation by 1.5 to match coefficients of \(a\):\[ 6a - 9c = 1.5 \]Now align this with the second:1.5. \( 6a - 9c = 1.5 \)2. \( 6a - 8c = 1 \)Subtract equation 2 from equation 1.5:\[ -9c + 8c = 1.5 - 1 \]\[ -c = 0.5 \]\[ c = -0.5 \].

Step 6: Calculate b and a

Using \(c = -0.5\), substitute into \(b = 2c\):\[ b = 2(-0.5) = -1 \].Now substitute \(c = -0.5\) back into \(4a - 6c = 1\) to solve for \(a\):\[ 4a - 6(-0.5) = 1 \]\[ 4a + 3 = 1 \]\[ 4a = -2 \]\[ a = -0.5 \].

Step 7: Verify Solution

We check the solution \((a, b, c) = (-0.5, -1, -0.5)\) in the original system.1. \(4(-0.5) - 3(-1) = -2 + 3 = 1\) ✔️2. \(6(-0.5) - 8(-0.5) = -3 + 4 = 1\) ✔️3. \(2(-1) - 4(-0.5) = -2 + 2 = 0\) ✔️All equations are satisfied, confirming our solution.

Key Concepts

Substitution Method

The substitution method is a technique used to solve systems of equations. It involves expressing one variable in terms of another and using this expression to substitute into the other equations.

This method is particularly useful when one equation is easily solvable for one of the variables. For instance, in our original exercise, we used the third equation, \(2b - 4c = 0\), to express \(b\) as \(b = 2c\). By substituting \(b = 2c\) into the first equation, we simplified the system and reduced the complexity, making the equations easier to handle.

When applying the substitution method:

• Choose an equation where a variable can be easily isolated.
• Solve for the chosen variable.
• Substitute this expression into the other equations, reducing the number of variables.
• Solve the resulting simpler system.
• Back-substitute to find the remaining variables, if needed.

It's a systematic method and, in many cases, it simplifies the problem significantly.

Elimination Method

The elimination method is another technique to solve systems of equations, especially useful for systems difficult to manage through substitution.

This method focuses on adding or subtracting equations to eliminate a variable entirely. In our exercise, once we had the reduced system, we aimed to eliminate a variable by manipulating the equations. For this, we multiplied the first reduced equation, \(4a - 6c = 1\), by 1.5 to match the coefficient of \(a\) in the second equation, \(6a - 8c = 1\).

Steps to use the elimination method:

• Align equations such that adding or subtracting them cancels out one variable.
• Rearrange or multiply equations to create equal coefficients for the variable you want to eliminate.
• Subtract or add the equations to eliminate the variable, leaving one equation with fewer variables.
• Solve the resulting simpler equation.
• Back-solve for other variables, if necessary.

This method is particularly effective when dealing with linear equations, as it quickly reduces the total number of unknowns.

Inconsistent Systems

An inconsistent system of equations is one where no solution exists that simultaneously satisfies all equations. This occurs when the equations represent parallel lines that never intersect.

Identifying an inconsistent system comes during the process of solving. While solving, any contradiction such as a false statement (like \(0 = 5\)) will indicate that no solutions exist.

In our exercise, all equations have valid solutions, meaning they intersect at a single point, confirming system consistency. However, understanding inconsistency is crucial for complex systems:

• Equations with identical slopes and different intercepts suggest parallel lines and an inconsistent system.
• Simplifying a system to a senseless equation confirms the system's inconsistency.
• Understanding inconsistent systems helps in modeling real-world problems where not all scenarios have solutions.

Learning to spot inconsistencies sharpens problem-solving skills and aids in understanding the geometry of equations.