Problem 34 - Algebra 2

Question

What is the solution to the system of equations shown below? \(\left\\{\begin{array}{l}{x-y+z=0} \\ {-5 x+3 y-2 z=-1} \\ {2 x-y+4 z=11}\end{array}\right.\) F. (0, 3, 3) G. (2, 5, 3) H. no solution J. infinitely many solutions

Step-by-Step Solution

Answer

The solution to the system of equations is (2, 5, 3), option G.

Step 1: Write the System of Equations

We are given the following system of equations to solve: \[ \begin{align*} x - y + z &= 0 \ -5x + 3y - 2z &= -1 \ 2x - y + 4z &= 11 \end{align*} \] We will use substitution or elimination to find the values of \(x\), \(y\), and \(z\).

Step 2: Simplify the First Equation

From the first equation \(x - y + z = 0\), we can express \(x\) in terms of \(y\) and \(z\): \[ x = y - z \] We will use this expression to substitute \(x\) in the other equations.

Step 3: Substitute x in Second and Third Equation

Substitute \(x = y - z\) into the second equation: \[ -5(y - z) + 3y - 2z = -1 \] Simplifying gives: \[ -5y + 5z + 3y - 2z = -1 \] Combine like terms: \[ -2y + 3z = -1 \]

Step 4: Substitute x in the Third Equation

Substitute \(x = y - z\) in the third equation: \[ 2(y - z) - y + 4z = 11 \] Simplifying gives: \[ 2y - 2z - y + 4z = 11 \] Combine like terms: \[ y + 2z = 11 \]

Step 5: Solve the Substituted System

We now have the following system of two equations: \[ \begin{align*} -2y + 3z &= -1 \ y + 2z &= 11 \end{align*} \] Multiply the second equation by 2: \[ 2y + 4z = 22 \]

Step 6: Eliminate y and Solve for z

Add the simplified equations: \[ \begin{align*} -2y + 3z &= -1 \ 2y + 4z &= 22 \end{align*} \] Add these equations: \[ 0y + 7z = 21 \] Solve for \(z\): \[ z = 3 \]

Step 7: Solve for y using z

Substitute \(z = 3\) back into one of the simplified equations, e.g., \(y + 2z = 11\): \[ y + 2(3) = 11 \] \[ y + 6 = 11 \] Solve for \(y\): \[ y = 5 \]

Step 8: Solve for x using y and z

Substitute \(y = 5\) and \(z = 3\) into the expression for \(x\): \[ x = y - z \] \[ x = 5 - 3 \] Solve for \(x\): \[ x = 2 \]

Step 9: Provide the Solution

The solution to the system of equations is the ordered triple \((x, y, z) = (2, 5, 3)\).

Key Concepts

substitution method

The substitution method is a powerful technique for solving systems of equations, especially when you are dealing with linear equations. This method involves expressing one variable in terms of others by rearranging one of the given equations. In this exercise, we have a system with three variables: \(x\), \(y\), and \(z\).
To utilize substitution, we first chose the equation \(x - y + z = 0\) to express \(x\) as \(x = y - z\). This expression is then substituted into the other two equations to eliminate \(x\).
By doing this, the system simplifies from three equations to two, reducing the complexity and making it easier to solve.
**Advantages of the Substitution Method:**

• It simplifies the problem by reducing the number of variables in each equation.
• It's generally straightforward for systems with one equation that is easy to solve for a single variable.

After applying these substitutions, you derive a simpler system that you can solve using arithmetic methods.

elimination method

The elimination method is another approach to solve systems of equations. This technique is often used when it's tricky to use substitution directly or when dealing with multiple variables.
The main idea behind elimination is to add or subtract equations to cancel out one or more of the variables, thus reducing the system's dimensionality.
**In Practice:**
From the exercise provided, after substitution, we derived the two new equations:

• \(-2y + 3z = -1\)
• \(y + 2z = 11\)

We chose to eliminate \(y\) by multiplying the second equation by 2, which transformed it into \(2y + 4z = 22\). By adding this to the first equation \(-2y + 3z = -1\), \(y\) was eliminated, and the equation \(7z = 21\) was obtained.
**Pros of the Elimination Method:**

• It's effective for systems where the coefficients of a variable are easier to manipulate for cancellation.
• Can quickly simplify a system if certain coefficients already allow for easy elimination.

The elimination method complements substitution and is an excellent tool for efficiently handling multi-variable linear systems.

linear equations

Linear equations are equations where each term is either a constant or the product of a constant and a single variable. These equations represent straight lines when graphed and are fundamental in algebra.
In the system given, all equations are linear. They are written in the form: \( ax + by + cz = d \). Linear systems can range from very simple to complex, depending on the number of variables and equations.
**Key Characteristic of Linear Systems:**

• Linear equations have solutions that can be graphically interpreted as the point(s) of intersection of the lines or planes represented by the equations.
• Understanding the behavior and solutions of these systems provides insight into more complex algebraic concepts.

In this exercise, solving the given linear system allowed us to find the precise values for \(x\), \(y\), and \(z\), ensuring they satisfy all three equations simultaneously.