Problem 30 - Precalculus with Limits

Question

Solving a System of Linear Equations In Exercises \(25 - 46\) , solve the system of linear equations and check any solutions algebraically. $$\left\\{ \begin{aligned} 5 x - 3 y + 2 z & = 3 \\ 2 x + 4 y - z & = 7 \\ x - 11 y + 4 z & = 3 \end{aligned} \right.$$

Step-by-Step Solution

Answer

The solution to the system of equations is \( x = -5, y = 1, z = -1 \).

Step 1: Setting up the Equations

We start by setting the system of equations as follows: \[5x - 3y + 2z = 3\] \[2x + 4y - z = 7\] \[x - 11y + 4z = 3\]

Step 2: Solving for x

Multiply second equation by 2 and subtract it from the first to eliminate y and z. This will give us the equation for x. \[(5x - 3y + 2z) - 2(2x + 4y - z) = 3 - 2 * 7\] The result is \[x = -5\]

Step 3: Solving for y and z

Substitute x = -5 into first and second equations to solve for y and z respectively. From first equation we find \(y = 1\) and from second equation we find \(z = -1\)

Step 4: Checking the solution

Now that we got the solutions, insert these values into the third equation and check whether the left-hand side equals the right-hand side to verify if the solutions are correct. After inserting \( (-5, 1, -1)\) into third equation we find that \( -5 -11*1 + 4*-1\) is indeed equal to 3.

Key Concepts

Algebraic Solution

An algebraic solution involves using algebraic methods to solve a system of equations. In the given problem, we solve a system of linear equations with three variables:

• 5x - 3y + 2z = 3
• 2x + 4y - z = 7
• x - 11y + 4z = 3

Algebraic solutions provide a concrete answer by manipulating these equations using mathematical operations. The main goal is to break down complex systems into simpler equations that can be solved individually. Successive elimination or substitution helps isolate each variable step by step. Once all variables are found, these values make up the complete solution to the system. By doing this algebraically, we ensure that the accuracy and consistency of the solution are well-maintained across all equations.

Substitution Method

The substitution method is one way to solve systems of equations. It involves expressing one variable in terms of others from one equation, and substituting this expression into another equation. In this exercise, to find each variable, a sequence of substitution steps was used. The steps began by manipulating the given equations so that all terms except one become nullified in one equation, allowing for a straightforward calculation of one variable. For example, the initial change was to adjust the equations to remove y and z by multiplying and subtracting equations, resulting in finding that x = -5.
Once x = -5 was known, this value was substituted back into the remaining equations, making it simpler to isolate and solve for y and z. Substituting the known values back helps in further simplification and solving for the remaining variables one at a time until all are determined.

Equation Verification

Equation verification is crucial to confirm that the obtained solutions satisfy all given equations. After solving the system in our exercise, the last step involved verifying the solution by plugging the values (-5, 1, -1) back into the equations. By inserting these values into the third equation, substituting each variable, we verified whether the left side of the equation equaled the right side, which was 3, ensuring the accuracy of the solution.

• This check eliminates any minor errors that may arise during calculations.
• It provides a clear proof that the solutions not only satisfy one equation, but all equations in the system.

Verification ensures the dependability of solutions in algebraic contexts.

Linear Algebra

Linear algebra is a branch of mathematics focused on vector spaces and linear mappings between them. It's foundational for solving systems of linear equations like the one given in the exercise. It provides frameworks such as matrices and determinants, which offer structured approaches to finding the solutions of systems. In this context, the problem used basic linear algebra principles, such as eliminating variables through linear combinations of equations. These principles simplify the complexity when systems involve multiple equations and variables.
Although the exercise focuses on elementary algebraic strategies, understanding linear algebra helps to appreciate how these methods scale up to more complex problems, like those involving larger systems of equations or higher dimensions in vector spaces.