Problem 14 - College Algebra with Corequisite Support

Question

For the following exercises, solve each system by elimination. $$ \begin{array}{l} 4 x-3 y+5 z=31 \\ -x+2 y+4 z=20 \\ x+5 y-2 z=-29 \end{array} $$

Step-by-Step Solution

Answer

\(x = -2\), \(y = -3\), \(z = 6\).

Step 1: Arrange Equations

We are given the system of equations: 1. \(4x - 3y + 5z = 31\)2. \(-x + 2y + 4z = 20\)3. \(x + 5y - 2z = -29\)Our goal is to use the elimination method to solve this system. First, let's work with equations 2 and 3 to eliminate one of the variables, such as \(x\).

Step 2: Eliminate x from Equations 2 and 3

To eliminate \(x\), we add equation 2 and equation 3:Equation 2: \(-x + 2y + 4z = 20\)Equation 3: \(x + 5y - 2z = -29\)Adding them gives:\((-x + x) + (2y + 5y) + (4z - 2z) = 20 - 29\)\(0x + 7y + 2z = -9\)This simplifies to: \(7y + 2z = -9\) (Equation 4).

Step 3: Eliminate x from Equations 1 and 2

Now, let's eliminate \(x\) by combining equations 1 and 2. To do this, multiply equation 2 by 4:Equation 2: \(-x + 2y + 4z = 20\) becomes \(-4x + 8y + 16z = 80\).Equation 1: \(4x - 3y + 5z = 31\)Add these results:\((4x - 4x) + (-3y + 8y) + (5z + 16z) = 31 + 80\)\(0x + 5y + 21z = 111\)This simplifies to: \(5y + 21z = 111\) (Equation 5).

Step 4: Solve for y and z

Now solve the two equations obtained from elimination (equation 4 and equation 5) with elimination:Equation 4: \(7y + 2z = -9\)Equation 5: \(5y + 21z = 111\)Multiply equation 4 by 5 and equation 5 by 7:\(35y + 10z = -45\)\(35y + 147z = 777\)Subtract:\((35y - 35y) + (10z - 147z) = -45 - 777\)\(-137z = -822\)Solve for \(z\):\(z = \frac{-822}{-137} = 6\).

Step 5: Substitute z back into Equation 4

Now, substitute \(z = 6\) back into equation 4:\(7y + 2(6) = -9\)This simplifies to:\(7y + 12 = -9\)Solve for \(y\):\(7y = -9 - 12\)\(7y = -21\)\(y = -3\).

Step 6: Solve for x using an Original Equation

Finally, substitute \(y = -3\) and \(z = 6\) back into one of the original equations to find \(x\). Use equation 3:\(x + 5(-3) - 2(6) = -29\)This simplifies to:\(x - 15 - 12 = -29\)\(x - 27 = -29\)Solve for \(x\):\(x = -29 + 27\)\(x = -2\).

Step 7: Conclusion

The solution to the system of equations is \(x = -2\), \(y = -3\), and \(z = 6\).

Key Concepts

Elimination Method

The elimination method is a powerful tool used in solving systems of linear equations. It revolves around the idea of removing, or "eliminating," variables to simplify the process of finding the set solution. By strategically adding or subtracting equations from each other, variables are systematically removed.

Here's how it works:

• Choose a variable to eliminate. In our case, we started with eliminating \(x\).
• Combine equations by addition or subtraction to remove the chosen variable. For example, adding equations may result in the coefficient of the chosen variable being zero.
• This step results in a new equation with one less variable, thereby simplifying the system.

This method often requires adjustments, such as multiplying an entire equation by a constant, to ensure the coefficients align for elimination. It is particularly effective when systems have variables that can be easily paired for elimination.

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, matrices, and linear equations. It forms the backbone for solving systems of equations like the one we tackled.

In linear algebra, systems of equations can be represented using matrices, with each equation forming a row of the matrix. This approach makes it easier to manipulate equations using matrix operations.

• Each row corresponds to an equation in the system.
• Each column represents a coefficient of a variable.

The concepts of linear algebra extend to operations like row reduction, which is analogous to our elimination step-by-step process. By understanding these operations, one can systematically solve complex systems, find solutions efficiently, and understand the properties of the systems themselves.

Linear algebra also introduces the concept of vectors and linear combinations, which are crucial in forming solutions for systems and understanding their graphical representations like planes and lines.

Solution of Equations

The solution of equations refers to finding values for the variables that satisfy all equations in a system simultaneously. In our exercise, we found the values \(x = -2\), \(y = -3\), and \(z = 6\) that solved each equation at the same time.

Here are steps often followed in finding such solutions:

• Use methods like substitution, elimination, or matrix operations to simplify the system.
• Solve the reduced system for one variable at a time ("back-substitution").
• Verify the solution by substituting back into the original equations to ensure all are satisfied.

Solving systems efficiently often requires strategic choices in methods and an understanding of equation properties such as linearity and dependence. The solution process not only determines if a solution exists but also reveals the nature of the solution—whether it is unique, infinite, or none at all.