Problem 29 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{aligned} x+y+z &=57 ,\\\ -2 x+y \quad \quad &=3 ,\\\ x- \quad \quad \quad& z=6 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( (x, y, z) = (15, 33, 9) \).

Step 1: Analyze the given system

The system of equations is: 1) \( x + y + z = 57 \)2) \( -2x + y = 3 \)3) \( x - z = 6 \)

Step 2: Isolate variables

First, isolate a variable. From equation 3, we can express \( z \) in terms of \( x \):\[ x - z = 6 \Rightarrow z = x - 6 \]

Step 3: Substitute \( z \) in the first equation

Substitute \( z = x - 6 \) into the first equation to get:\[ x + y + (x - 6) = 57 \Rightarrow 2x + y - 6 = 57 \Rightarrow 2x + y = 63 \]

Step 4: Solve the simplified system

Now we have a new system of equations:1) \( 2x + y = 63 \)2) \( -2x + y = 3 \)Subtract the second simplified equation from the first:\[ (2x + y) - (-2x + y) = 63 - 3 \Rightarrow 4x = 60 \Rightarrow x = 15 \]

Step 5: Find \( y \) and \( z \)

Use \( x = 15 \) in the second simplified equation to find \( y \):\[ -2(15) + y = 3 \Rightarrow -30 + y = 3 \Rightarrow y = 33 \]Next, use \( x = 15 \) in the isolated equation for \( z \):\[ z = x - 6 \Rightarrow z = 15 - 6 \Rightarrow z = 9 \]

Key Concepts

Solving Systems of Equations

When solving a system of equations, we aim to find the values of the variables that satisfy all equations simultaneously. In the given exercise, we have three variables: \(x\), \(y\), and \(z\). The system of equations is:

\[ \begin{aligned} x + y + z &=57 ,\ -2x + y \quad &=3 ,\ x - z &= 6 \end{aligned} \]

To find the solution, we will use a method called the Substitution Method. Let’s take a closer look at how this method works and its application in solving our system.

Substitution Method

The substitution method involves isolating one variable in one equation and substituting this expression into the other equations. This reduces the number of variables step-by-step until we can solve the system easily.

Step-by-Step Process:

• Step 1: Analyze the given system and choose an equation to isolate a variable.
• Step 2: Substitute the isolated variable into the other equations to simplify the system.
• Step 3: Solve the simplified system for the remaining variables.

In the given problem:
Equation 1: \(x + y + z = 57\)
Equation 2: \(-2x + y = 3\)
Equation 3: \(x - z = 6\)

We start by isolating \(z\) in Equation 3:
\(z = x - 6\).
Next, we substitute this expression into Equation 1:
\(x + y + (x - 6) = 57\), which simplifies to \(2x + y = 63\).

Now we have a new system of two equations and two variables:
1) \(2x + y = 63\)
2) \(-2x + y = 3\).
Subtracting the second equation from the first, we get:
\(4x = 60\), thus \(x = 15\).
Using this in the second equation, we find \(y\):
\(y = 33\). Lastly, substituting \(x = 15\) into the isolated equation for \(z\), we get:
\(z = 9\).

Linear Algebra

Linear algebra is a branch of mathematics dealing with vectors, vector spaces (or linear spaces), linear transformations, and systems of linear equations. When solving systems of linear equations, like in our problem, we aim to find values for the variables that satisfy all given equations simultaneously.

Understanding the basics of linear algebra helps in solving different types of linear equations efficiently. Here are some core aspects:

• Vector Spaces: Collection of vectors where vector addition and scalar multiplication are defined.
• Linear Equations: Equations that graph as straight lines or planes in multidimensional space.
• Matrix Operations: Tools to simplify and solve systems of linear equations, especially complex ones.

In our problem, we used basic algebra and the substitution method. Knowing linear algebra can extend these methods to more complex systems, making it easier to analyze and solve them with precision.