Question

Solve the system of linear equations. $$ \begin{aligned} x+y-z &=1 \\ 2 x+y &=2 \\ y+2 z &=0 \end{aligned} $$

Short answer

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

Step-by-step solution

Simplify the equations

We'll work with the first equation: \( x + y - z = 1 \). Rearrange it to express one variable in terms of the others: \( x = 1 - y + z \). This will help us substitute values later.

Express \( x \) from the second equation

Now, use the second equation to express \( x \) in terms of \( y \): \( 2x + y = 2 \) implies \( 2x = 2 - y \), thus \( x = 1 - \frac{y}{2} \).

Solve for \( y \) from the third equation

From the third equation, \( y + 2z = 0 \), express \( y \) in terms of \( z \): \( y = -2z \).

Substitute \( y \) into the expression for \( x \) in Step 2

Substitute \( y = -2z \) into \( x = 1 - \frac{y}{2} \), thus \( x = 1 - \frac{-2z}{2} \), simplifying to \( x = 1 + z \).

Substitute expressions for \( y \) and \( x \) into equation from Step 1

Now substitute \( x = 1 + z \) and \( y = -2z \) into the equation \( x + y - z = 1 \): \((1+z) + (-2z) - z = 1 \), simplifying gives \( 1 - 2z = 1 \).

Solve for \( z \)

The equation from the previous step \( 1 - 2z = 1 \) simplifies to \( -2z = 0 \), giving us \( z = 0 \).

Find \( y \) using \( z \)

Substitute \( z = 0 \) back into \( y = -2z \): \( y = -2(0) = 0 \).

Find \( x \) using \( z \)

Substitute \( z = 0 \) into \( x = 1 + z \): \( x = 1 + 0 = 1 \).

Key concepts

Linear Equations

Linear equations are algebraic expressions that represent straight lines when graphed on a coordinate plane. Each equation consists of variables, coefficients, and constants that can depict real-world situations.

When dealing with a system of linear equations like in this exercise, we're observing multiple equations working together. We aim to find the set of values for the variables that satisfy all the equations at once. This includes understanding their intersections on a graph, which represent the solution set.

In a typical system such as:

• \( x + y - z = 1 \)
• \( 2x + y = 2 \)
• \( y + 2z = 0 \)

These equations collectively mirror lines or planes in three-dimensional space, and solving them means pinpointing the exact spot where they all meet.

Substitution Method

The substitution method is a technique used to solve systems of equations. You begin by solving one of the equations for one of its variables.

In this exercise, we rearranged the first equation to express \( x \) in terms of \( y \) and \( z \):
\( x = 1 - y + z \).

This new expression of \( x \) becomes crucial later, allowing us to replace it in the other equations. By substituting into the second equation \( 2x + y = 2 \), we express \( x \) as:
\( x = 1 - \frac{y}{2} \)
This step aids in reducing the original system into more manageable parts, focusing on two variables for each equation until we isolate all variables.

The beauty of substitution lies in its simplicity; it reduces complexities incrementally while drilling down into the solution.

Solving Equations

Solving equations is the fundamental process of determining the unknowns that satisfy given equations. Each step helps isolate each variable, often using operations like addition, subtraction, multiplication, and division.

Take equation three for example, \( y + 2z = 0 \). This step isolates \( y \) as \( y = -2z \). With this equation, knowing the value of \( z \) directly gives us \( y \).

As you continue, you substitute these known values back into other expressions. For instance, \( z \), calculated as zero earlier, is replaced back into the expression for \( y \) to confirm its value as zero.

Solving equations glues the whole system together, confirming each variable's specific values and checking if all equations hold true simultaneously.

Matrix Algebra

Matrix algebra is a powerful tool for solving systems of linear equations, especially when dealing with multiple equations and variables. It involves arranging coefficients of the variables into a matrix form.

In matrix algebra, the system of equations:

• \( x + y - z = 1 \)
• \( 2x + y = 2 \)
• \( y + 2z = 0 \)

can be represented as a matrix equation, making use of rows and columns to simplify the manipulation of these equations.

The use of matrix operations, such as row reduction or finding the inverse of a matrix, provides a pathway to solve these equations efficiently. It is especially useful when you're dealing with larger systems or when computational tools are at hand.

Matrix algebra complements the substitution and elimination methods, offering a versatile approach to solving equations.