Problem 8 - Precalculus

Question

Solve each system of linear equations. $$\begin{array}{rr} x-2 y+z= & 0 \\ -2 x+y-z= & -5 \\ 13 x+7 y+5 z= & 6 \end{array}$$

Step-by-Step Solution

Answer

The solution is \( x = \frac{91}{25}, y = -\frac{34}{25}, z = -\frac{159}{25} \).

Step 1: Identify the system of equations

The given system of equations is: 1. \( x - 2y + z = 0 \)2. \( -2x + y - z = -5 \)3. \( 13x + 7y + 5z = 6 \)We need to solve for \( x \), \( y \), and \( z \).

Step 2: Solve for x and z in terms of y from equations 1 and 2

From equation 1, solve for \( z \):\[ z = 2y - x \]From equation 2, solve for \( z \) again:\[ z = -2x + y + 5 \]

Step 3: Set equations for z equal and solve for x in terms of y

Set the two expressions for \( z \) equal:\[ 2y - x = -2x + y + 5 \]Rearrange terms to solve for \( x \):\[ x = y + 5 \]

Step 4: Substitute x in terms of y back into equation for z

Substitute \( x = y + 5 \) into \( z = 2y - x \):\[ z = 2y - (y + 5) = y - 5 \]

Step 5: Substitute expressions for x and z into equation 3

Substitute \( x = y + 5 \) and \( z = y - 5 \) into equation 3:\[ 13(y + 5) + 7y + 5(y - 5) = 6 \]

Step 6: Simplify equation to solve for y

Expand and simplify the expression:\[ 13y + 65 + 7y + 5y - 25 = 6 \]Combine like terms:\[ 25y + 40 = 6 \]Solve for \( y \):\[ 25y = -34 \]\[ y = -\frac{34}{25} \]

Step 7: Substitute value of y to find x and z

Substitute \( y = -\frac{34}{25} \) into \( x = y + 5 \):\[ x = -\frac{34}{25} + 5 = \frac{91}{25} \]Substitute \( y = -\frac{34}{25} \) into \( z = y - 5 \):\[ z = -\frac{34}{25} - 5 = -\frac{159}{25} \]

Step 8: Verify the solution

Substitute \( x = \frac{91}{25} \), \( y = -\frac{34}{25} \), and \( z = -\frac{159}{25} \) back into the original equations to ensure all equations are satisfied.

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces (also known as linear spaces), linear transformations, and systems of linear equations.

• Basics: At its core, linear algebra revolves around the study of linear equations and their representations through matrices and vector spaces. A vector space is a collection of vectors that can be added together and multiplied by scalar numbers, following specific rules.
• Importance: Linear algebra is foundational for various mathematical calculations and practical applications, including computer science, physics, and engineering. It provides tools for modeling and solving problems where variables interact linearly, making it instrumental in fields like data science and machine learning.

By understanding linear algebra, you can effectively analyze systems of equations, like the one given in our exercise. It helps you see how multiple equations can intersect and provide a solution for unknown variables.

Substitution Method

The substitution method is a straightforward technique used to solve systems of equations. In this method, you take one of the equations and solve for one variable in terms of the others. This expression is then substituted into other equations to eventually find values for all unknowns.

• Step-by-Step Approach: Begin by choosing an equation and isolating one variable. Next, substitute this expression into the other equations in the system. Repeat the process until each variable is solved. Finally, substitute back to verify the solution.
• Advantages: It's particularly useful for systems with fewer equations or where one equation already clearly presents a variable. Substitution allows you to reduce the complexity of a system, making it easier to handle.

In the provided exercise, we use substitution by expressing variables from equations and substituting them into others, transforming the system into simpler terms that reveal the solution.

Solving Equations

Solving equations is about finding values for unknown variables that make equations true. In the context of systems of linear equations, this means finding a solution that satisfies all equations in the system simultaneously.

• Types of Solutions: Systems of linear equations can have a unique solution, infinitely many solutions, or no solution at all. This depends on the relationships and intersections of the equations.
• Consistency and Dependency: If a system graphically shows all lines intersecting at one point, it has a unique solution and is consistent. If the lines overlap, there are infinitely many solutions, indicating dependency. If they do not intersect, the system has no solution and is inconsistent.

In the exercise, finding the solutions for \(x\), \(y\), and \(z\) confirms the system’s consistency, meaning values were found that satisfy all the given equations. This outcome is what solving systems of linear equations aims to achieve.