Problem 16 - Algebra and Trigonometry

Question

\(13-16\) . Perform an operation on the given system that eliminates the indicated variable. Write the new equivalent system. $$ \begin{array}{c}{x-4 y+z=3} \\ {y-3 z=10} \\ {3 y-8 z=24} \\ {\text { Eliminate the } y \text { -term }} \\ {\text { from the third equation. }}\end{array} $$

Step-by-Step Solution

Answer

The new system is: \(x - 4y + z = 3\), \(y - 3z = 10\), \(z = -6\).

Step 1: Identify the Equation to Modify

We are given three equations, and we need to eliminate the \(y\)-term from the third equation, which is \(3y - 8z = 24\). To do this, we will use the second equation \(y - 3z = 10\).

Step 2: Solve for \( y \) in the Second Equation

First, solve the second equation for \(y\):\[ y = 3z + 10 \]

Step 3: Substitute \( y \) in the Third Equation

Replace \(y\) in the third equation \(3y - 8z = 24\) with the expression found in Step 2: \[ 3(3z + 10) - 8z = 24 \]

Step 4: Simplify the Equation

Expand and simplify the substituted equation: \[ 9z + 30 - 8z = 24 \]. This simplifies to \[ z + 30 = 24 \].

Step 5: Solve for \( z \)

Subtract 30 from both sides to solve for \(z\): \[ z = 24 - 30 \] which simplifies to \[ z = -6 \].

Step 6: Write the New Equivalent System

The new system of equations is as follows: \(x - 4y + z = 3\), \(y - 3z = 10\), and the modified third equation is \(z = -6\).

Key Concepts

Elimination Method

The elimination method is a popular technique for solving systems of equations. It involves eliminating variables sequentially to make the system easier to solve. This method can be especially useful when dealing with multiple linear equations. Here's how it works in simple steps:

• Identify the variable you want to eliminate and choose two equations where this variable is present.
• Multiply the equations by appropriate coefficients so that when they are added, the chosen variable cancels out.
• Add or subtract the equations to eliminate the selected variable.

In the given exercise, the goal was to eliminate the y-term from the third equation. By substituting from the second equation after simplifying, we managed to isolate another variable, making the system more manageable.

Substitution Method

The substitution method is another powerful tool for solving systems of equations. It involves solving one equation for one variable and then substituting the result into another equation.This is a great choice when one equation can be easily solved for a variable with minimal manipulation. Here are the basic steps:

• Solve one of the equations for one of its variables.
• Substitute this expression into the other equations in the system.
• Solve the resulting single-variable equation.

In our exercise, we first solved the second equation for y: \[ y = 3z + 10 \]We then substituted this expression into the third equation, helping us to focus entirely on scrutinizing the z-terms and make further progress towards solving the system.

Linear Equations

Linear equations are equations where the variables are raised to the power of one. They form the backbone of many algebraic problems and can represent anything from simple arithmetic to complex systems. These equations can be solved using various techniques like the ones we discussed (elimination and substitution). Some key characteristics include:

• The graph of a linear equation in two variables is a straight line.
• They have constant rates of change and are predictable.
• They often come in a standard form: Ax + By = C, where A, B, and C are constants.

In our system, the original equations are all linear. By using methods such as elimination or substitution, we can transform these equations, reduce the number of variables to be dealt with, and find solutions efficiently.