Problem 31 - Intermediate Algebra

Question

Give a step-by-step description of how to solve the system $$ \left(\begin{array}{rr} x-2 y+3 z= & -23 \\ 5 y-2 z= & 32 \\ 4 z= & -24 \end{array}\right) $$

Step-by-Step Solution

Answer

The solution is \(x = 3\), \(y = 4\), \(z = -6\).

Step 1: Solve the third equation for z

The third equation in the system is \(4z = -24\). To find \(z\), divide both sides of the equation by 4. \[ z = \frac{-24}{4} = -6 \] So, \(z = -6\).

Step 2: Substitute z into the second equation to find y

With \(z = -6\) found from Step 1, substitute \(z\) into the second equation: \[5y - 2(-6) = 32\] Simplify the equation:\[5y + 12 = 32\] Subtract 12 from both sides:\[5y = 20\] Divide by 5:\[ y = \frac{20}{5} = 4\] So, \(y = 4\).

Step 3: Substitute y and z into the first equation to find x

Now substitute \(y = 4\) and \(z = -6\) into the first equation:\[ x - 2(4) + 3(-6) = -23\] Simplify:\[ x - 8 - 18 = -23\] Combine like terms:\[ x - 26 = -23\] Add 26 to both sides:\[ x = -23 + 26\] So, \(x = 3\).

Step 4: Verify the solution

Substitute \(x = 3\), \(y = 4\), and \(z = -6\) back into each original equation to ensure they are satisfied.For the first equation:\[ 3 - 2(4) + 3(-6) = 3 - 8 - 18 = -23 \] For the second equation:\[ 5(4) - 2(-6) = 20 + 12 = 32 \] For the third equation:\[ 4(-6) = -24 \] All equations hold true with these values, confirming the solution is correct.

Key Concepts

Linear Algebra

Linear algebra is the branch of mathematics that deals with systems of equations and the properties of vectors and matrices. It provides the foundation for solving equations like those encountered in the given system. A key part of linear algebra involves representing equations as matrices or vectors.
Imagine equations as different lanes in a highway where each lane can intersect with another at designated points. These intersections help us find solutions. When handling system of equations, linear algebra helps break down complex networks of interrelated variables. It lets us simplify problems, making it easier to solve them systematically. The key terms include vectors, which are like arrows specifying direction and magnitude, and matrices, which are rectangular arrangements of numbers. These structures help organize information and find solutions efficiently.
In the exercise we are working with, there's a system of three linear equations. Solving these using linear algebra techniques, such as the substitution method outlined in our example, helps us find specific values for each variable.

Substitution Method

The substitution method is a powerful technique used in solving systems of equations. It involves solving one equation for a particular variable and then substituting this into another equation. This transforms the complex system into a simpler one-variable equation, which is easier to solve.
To break it down:

• Solve one of the equations for one of the variables.
• Substitute this expression into another equation, replacing the variable.
• Repeat if necessary and solve the resulting simpler equation.
• Continue until all variables are determined.

In our step-by-step solution, we started by isolating \( z \) in the third equation. This allowed us to substitute \( z = -6 \) into the second equation, simplifying it to find \( y = 4 \). Finally, knowing \( y \) and \( z \), we substituted both into the first equation to find \( x = 3 \).
This method is excellent when one equation is easily solvable for a single variable, especially when equations contain two or more unknowns. It systematically untangles the equations, making complex systems manageable.

Verification of Solutions

Verification is a crucial step in solving systems of equations, as it ensures the solutions obtained are indeed correct. After finding potential solutions, you substitute them back into the original equations to check that each equation holds true.
This process is like checking your work to ensure there were no errors along the way:

• Take each solution and plug it back into the original equations.
• Go through each equation to see if both sides are equal after substitution.
• If they are, the solution is verified. If not, re-evaluate your steps to find errors.

In our example, substituting \( x = 3 \), \( y = 4 \), and \( z = -6 \) into each of the original equations showed they satisfy all equations. This means the solutions are indeed correct, confirming our steps were accurate. Verification is particularly important in mathematics as it reinforces understanding and provides confidence in results.