Problem 36 - Elementary and Intermediate Algebra

Question

Solve each system. If a system’s equations are dependent or if there is no solution, state this. $$\begin{aligned} x+y \quad \quad =0 ,\\\ x \quad \quad+z=1 ,\\\ 2 x+y+z =2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(x = 1, y = -1, z = 0\).

Step 1 - Set Up the Equations

Start with the equations given in the problem:1. \(x + y = 0\)2. \(x + z = 1\)3. \(2x + y + z = 2\)

Step 2 - Express y from the First Equation

From the first equation \(x + y = 0\), solve for \(y\):\(y = -x\)

Step 3 - Express z from the Second Equation

From the second equation \(x + z = 1\), solve for \(z\):\(z = 1 - x\)

Step 4 - Substitute y and z into the Third Equation

Substitute \(y = -x\) and \(z = 1 - x\) into the third equation \(2x + y + z = 2\):\(2x + (-x) + (1 - x) = 2\)Simplify the equation:\(2x - x + 1 - x = 2\)Combine like terms:\(x + 1 = 2\)

Step 5 - Solve for x

Solve \(x + 1 = 2\) for \(x\):\(x = 1\)

Step 6 - Find y and z

Using \(x = 1\), substitute back to find \(y\) and \(z\):\(y = -x = -1\)\(z = 1 - x = 0\)

Step 7 - Verify the Solution

Check the solution \(x = 1, y = -1, z = 0\) in all original equations:1. \(1 + (-1) = 0\) (True)2. \(1 + 0 = 1\) (True)3. \(2(1) + (-1) + 0 = 2\) (True)

Key Concepts

System of Equations

A system of equations is a collection of two or more equations involving the same set of variables. In our exercise, we are working with a system that includes three equations and three variables:

1. \(x + y = 0\)
2. \(x + z = 1\)
3. \(2x + y + z = 2\)

The objective is to find values for the variables \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously. Working with systems of equations allows us to model real-world problems where multiple conditions must be satisfied at the same time.

Linear Algebra

Linear algebra provides the framework for expressing and solving systems of equations efficiently. One of the fundamental tools in linear algebra is solving systems of linear equations using methods like substitution, elimination, and matrix operations.

Linear algebra helps us understand relationships between variables in multi-dimensional space. In our case, it aids in visualizing how the variables \(x\), \(y\), and \(z\) interact to satisfy the given equations. It also introduces advanced concepts like vector spaces and linear transformations, but those are topics for more advanced studies.

Substitution Method

The substitution method is one of the ways to solve systems of equations. The basic idea is to solve one of the equations for one variable and then substitute that expression into the other equations. This reduces the number of variables, making the equation simpler to solve step-by-step.

In our exercise, we used the substitution method as follows:
• From the first equation \(x + y = 0\), we solved for \(y\): \(y = -x\).
• From the second equation \(x + z = 1\), we solved for \(z\): \(z = 1 - x\).
• We then substituted \(y = -x\) and \(z = 1 - x\) into the third equation \(2x + y + z = 2\) to find \(x\).

This substitution reduced the problem to a single variable equation that was easier to solve.

Dependent Equations

Dependent equations are equations that describe the same line or plane in geometry, meaning they have the exact same set of solutions. In the context of linear systems, dependent equations can result in infinitely many solutions or indicate redundancy in the system.

In our exercise, the equations given were not dependent since each equation provided unique information to determine the values of \(x\), \(y\), and \(z\). However, recognizing dependent equations can simplify the problem-solving process, as it reduces the number of unique equations we need to work with.

Solution Verification

Once you have found the values for the variables that satisfy the system of equations, it is crucial to verify your solution. You can do this by substituting the values back into the original equations to ensure they hold true.

In our exercise, we found:
• \(x = 1\)
• \(y = -1\)
• \(z = 0\)

Let's verify these by substituting them back into each original equation:

• For the first equation \(x + y = 0\): \(1 - 1 = 0\), which is true.
• For the second equation \(x + z = 1\): \(1 + 0 = 1\), which is true.
• For the third equation \(2x + y + z = 2\): \(2(1) - 1 + 0 = 2\), which is true.

Since all the equations hold true, our solution \(x = 1\), \(y = -1\), and \(z = 0\) is correct.