Problem 8 - Beginning and Intermediate Algebra

Question

Solve the system of equations. $$\begin{aligned} x+2 y-z &=4 \\ 4 x-3 y+z &=8 \\ 5 x-y &=12 \end{aligned}$$

Step-by-Step Solution

Answer

The general solution of the given system of equations is: \( \begin{aligned} x &= x \\\\ y &= 5x - 12 \\\\ z &= 11x - 28 \end{aligned} \) This represents an infinite number of solutions since for any value of x, there will be corresponding unique values for y and z.

Step 1: Determine which method to use

In this case, we will use the elimination method to make the equation solving simple.

Step 2: Eliminate one variable from two equations

Let's eliminate z from the first two equations. To do that, we can add the first equation and the second equation, since they have the opposite coefficients for z: \( \begin{aligned} (x+2y-z) + (4x-3y+z) &= 4+8 \end{aligned} \) Now simplify: \( \begin{aligned} 5x - y &= 12 \end{aligned} \) We are left with a new equation, which is actually the same as the third given equation.

Step 3: Observe that we have a dependent system

Notice that we acquired the same equation in step 2 as the given third equation. That means that we have a dependent system, and there are an infinite number of solutions. Now we need to express the solution in terms of one variable.

Step 4: Express y in terms of x

We already have the equation \(5x - y = 12\). We can rearrange it to solve for y: \( \begin{aligned} y = 5x - 12 \end{aligned} \)

Step 5: Substitute y in terms of x into the first equation

Now that we have y in terms of x, we can substitute it into the first equation to find z in terms of x: \( \begin{aligned} x + 2(5x - 12) - z &= 4 \end{aligned} \) Expand and simplify: \( \begin{aligned} x + 10x - 24 - z &= 4 \\\\ 11x - z &= 28 \end{aligned} \)

Step 6: Express z in terms of x

From the equation obtained in Step 5, we can express z in terms of x: \( \begin{aligned} z = 11x - 28 \end{aligned} \)

Step 7: State the general solution

Finally, we can present the general solution of the system in terms of the variable x: \( \begin{aligned} x &= x \\\\ y &= 5x - 12 \\\\ z &= 11x - 28 \end{aligned} \) For any value of x, there will be a unique solution for y and z. This means that the system has an infinite number of solutions.

Key Concepts

Elimination Method

The elimination method is a technique used to solve systems of linear equations. Instead of solving the equations directly for one variable at a time, this method involves combining equations to cancel out or eliminate one variable. This is accomplished by adding or subtracting the equations after multiplying them, if needed, to make the coefficients of one variable equal and opposite for cancellation.

In the exercise, we have three equations:

• \(x+2y-z = 4\)
• \(4x-3y+z = 8\)
• \(5x-y = 12\)

We eliminate the variable \(z\) by adding the first two equations which have opposite coefficients of \(z\). This simplifies the equation, helping us progress to the solution by reducing the number of variables.

Dependent System

A dependent system of equations occurs when all the equations represent the same geometric object, such as a line or a plane, in the graph. Essentially, the equations are related in such a way that they do not provide new information.

In the exercise, when the elimination method is used, the resultant equation \(5x - y = 12\) is identical to the third given equation. This indicates that the system of equations is dependent. In a dependent system, solving the equations transforms them into equivalent expressions that do not increase our understanding of the variables beyond what is initially provided.

Infinite Solutions

In a dependent system, the system of equations does not have a unique solution. Instead, there are infinite solutions. This means there are countless sets of values for the variables that satisfy all the given equations.

Since the system turned out to be dependent, every solution is a combination of values for \(x\), \(y\), and \(z\) derived from the general form we obtained:

• \(x = x\)
• \(y = 5x - 12\)
• \(z = 11x - 28\)

For any value of \(x\), you can find corresponding \(y\) and \(z\) values that will satisfy all the equations, illustrating that there are indeed infinite solutions available.

Expressing Variables in Terms of Another

Expressing one variable in terms of another is an effective strategy for solving systems of equations, especially with dependent systems. This allows us to represent the solution set in a simplified form.

Here, we started by expressing \(y\) in terms of \(x\) from the equation \(5x - y = 12\), which results in the expression \(y = 5x - 12\).

Next, this expression for \(y\) was substituted back into the first equation to solve for \(z\) in terms of \(x\), yielding \(z = 11x - 28\).

This transforms the problem into expressing solutions dependent solely on \(x\), making it easier to understand the relationship between the variables and see clearly how many solutions are available.