Problem 16 - Intermediate Algebra

Question

Solve each of the following systems. If the solution set is \(\varnothing\) or if it contains infinitely many solutions, then so indicate. $$ \left(\begin{array}{rr} 5 x-3 y+z= & 1 \\ 2 x-5 y= & -2 \\ 3 x-2 y-4 z= & -27 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \( x = 1 \), \( y = \frac{4}{5} \), \( z = -1.6 \).

Step 1: Analyze the System of Equations

We have three equations in three variables (x, y, z). We'll refer to them as Equation 1 \( 5x - 3y + z = 1 \), Equation 2 \( 2x - 5y = -2 \), and Equation 3 \( 3x - 2y - 4z = -27 \). Our goal is to solve for the values of \( x \), \( y \), and \( z \).

Step 2: Express z from Equation 1

Start with Equation 1: \( 5x - 3y + z = 1 \). Express \( z \) in terms of \( x \) and \( y \):\[ z = 1 - 5x + 3y \]

Step 3: Substitute z in Equation 3

Substitute the expression for \( z \) from Step 2 into Equation 3: \( 3x - 2y - 4(1 - 5x + 3y) = -27 \).Simplify it to get one equation in terms of \( x \) and \( y \):\[ 3x - 2y - 4 + 20x - 12y = -27 \] \\[ 23x - 14y = -23 \]

Step 4: Solve Equations 2 and the New Equation

Now, solve the system consisting of Equation 2 \( 2x - 5y = -2 \) and the new equation \( 23x - 14y = -23 \). Multiply Equation 2 by 7 to align coefficients:\[ 14x - 35y = -14 \] Subtract this from \( 23x - 14y = -23 \):\[ 9x + 21y = -9 \]Solve for \( x \):\[ x = 1 \]

Step 5: Find y Using Equation 2

Substitute \( x = 1 \) back into Equation 2 \( 2x - 5y = -2 \):\[ 2(1) - 5y = -2 \] \[ 2 - 5y = -2 \] \[ 5y = 4 \]\[ y = \frac{4}{5} \]

Step 6: Determine z Using x and y

Substitute \( x = 1 \) and \( y = \frac{4}{5} \) back into the expression for \( z \) from Step 2:\[ z = 1 - 5(1) + 3(\frac{4}{5}) \]\[ z = 1 - 5 + \frac{12}{5} \]\[ z = -4 + \frac{12}{5} = -4 + 2.4 = -1.6 \]

Step 7: Verify Solution

Verify the solution \((x, y, z) = (1, \frac{4}{5}, -1.6)\) by plugging it back into the original equations to ensure all hold true.- Equation 1: \( 5(1) - 3(\frac{4}{5}) - 1.6 = 1 \), holds true.- Equation 2: \( 2(1) - 5(\frac{4}{5}) = -2 \), holds true.- Equation 3: \( 3(1) - 2(\frac{4}{5}) - 4(-1.6) = -27 \), holds true.

Key Concepts

Linear Equations

A system of linear equations is a collection of two or more linear equations involving the same set of variables. Linear equations are algebraic equations in which each term is either a constant or the product of a constant and a single variable. In simpler terms, they form a straight line when graphed.

For example, consider the following system:

• Equation 1: \( 5x - 3y + z = 1 \)
• Equation 2: \( 2x - 5y = -2 \)
• Equation 3: \( 3x - 2y - 4z = -27 \)

Each equation in this system is linear because the variables \(x\), \(y\), and \(z\) are raised to the power of one and are not multiplied by each other.

The goal is to find a set of values for these variables that satisfies all equations simultaneously. Understanding how to solve such systems is vital in algebra and is widely used in various fields such as physics, engineering, and economics.

Elimination Method

The elimination method is a technique for solving a system of equations. The main goal of this method is to eliminate one of the variables by adding or subtracting the equations. This reduces the system to an equation with fewer unknowns, making it easier to solve.

In our exercise, we use the elimination method when we align the coefficients of the variable \(y\) in Equations 2 and the newly formed equation:

• Equation 2: \(2x - 5y = -2\)
• New Equation: \(23x - 14y = -23\)

By multiplying Equation 2 by 7, we get \(14x - 35y = -14\), which allows us to subtract it from the new equation. This subtraction yields the equation \(9x + 21y = -9\).

Now there's only one equation in terms of \(x\) and \(y\), which makes finding the solutions much more manageable. Through this method, elimination simplifies solving for one variable at a time.

Variable Substitution

Variable substitution involves solving one of the equations for one variable and substituting this expression into other equations. This reduces the number of variables in equations, simplifying the solution process.

In our exercise, we start by expressing \(z\) from Equation 1 as:

• \(z = 1 - 5x + 3y\)

Then, we substitute this expression into Equation 3:

• Original Equation 3: \(3x - 2y - 4z = -27\)
• Substituted Equation: \(3x - 2y - 4(1 - 5x + 3y) = -27\)

This substitution allows us to eliminate \(z\) from Equation 3, resulting in an equation with only \(x\) and \(y\). The substitution method efficiently reduces the complexity of solving a system of equations by reducing it to a simpler one with fewer unknowns.

Solution Verification

Solution verification is key to ensuring the correctness of the solution obtained from a system of equations. Once you determine the values, you should substitute them back into the original equations to verify that they satisfy each equation.

For the solution \((x, y, z) = (1, \frac{4}{5}, -1.6)\), we check:

• Equation 1: \(5(1) - 3(\frac{4}{5}) - 1.6 = 1\) holds true.
• Equation 2: \(2(1) - 5(\frac{4}{5}) = -2\) holds true.
• Equation 3: \(3(1) - 2(\frac{4}{5}) - 4(-1.6) = -27\) holds true.

Each equation confirms the solution satisfies the system. This process ensures any errors in calculation are caught, providing confidence in the solutions, and demonstrating a complete understanding.