Problem 53 - Elementary and Intermediate Algebra

Question

For Exercises 52 and \(53,\) let u represent \(1 / x,\) v represent \(1 / y,\) and \(w\) represent \(1 / z .\) Solve for \(u, v,\) and \(w,\) and then solve for \(x, y,\) and \(z\). $$\begin{aligned} &\frac{2}{x}+\frac{2}{y}-\frac{3}{z}=3,\\\ &\frac{1}{x}-\frac{2}{y}-\frac{3}{z}=9,\\\ &\frac{7}{x}-\frac{2}{y}+\frac{9}{z}=-39 \end{aligned}$$

Step-by-Step Solution

Answer

Solve the linear system for u, v, and w, then invert these to get x, y, and z.

Step 1: Substitute u, v, and w

First, substitute the variables given: let u = \(\frac{1}{x}\), v = \(\frac{1}{y}\), and w = \(\frac{1}{z}\). Rewrite the system of equations in terms of u, v, and w: \[2u + 2v - 3w = 3\] \[u - 2v - 3w = 9\] \[7u - 2v + 9w = -39\]

Step 2: Solve for u, v, and w

Solve the system of linear equations using matrix methods, substitution, or elimination. Let's use the elimination method for simplicity.First combine the equations to eliminate one of the variables. Subtract the second equation from the first to eliminate v: \[(2u + 2v - 3w) - (u - 2v - 3w) = 3 - 9\] \[u + 4v = -6\]Next, solve for u: \[u = -6 - 4v\]

Step 3: Eliminate another variable

We can now use another equation to eliminate w. Multiply the second equation by 2 and subtract the third equation from it: \[2(u - 2v - 3w) - (7u - 2v + 9w) = 2(9) - (-39)\] \[2u - 4v - 6w - 7u + 2v - 9w = 18 + 39\] \[-5u - 2v - 15w = 57\]

Step 4: Solve the reduced system

Substitute \(u = -6 - 4v\) into the reduced system: \[-5(-6 - 4v) - 2v - 15w = 57\] \[30 + 20v - 2v - 15w = 57\] \[30 + 18v - 15w = 57\] \[18v - 15w = 27\]Solve for v: \[v = \frac{27 + 15w}{18}\]

Step 5: Solve for w

Now substitute v back into \(u = -6 - 4v\) and solve for w. Using any original equation, substitute these results back to find w. Once w is found, use it to find v and then u using substitution.

Step 6: Solve for x, y, and z

After determining the values of u, v, and w from solving the system, revert to the original variables: \(x = \frac{1}{u}\) \(y = \frac{1}{v}\) \(z = \frac{1}{w}\)Calculate these values to find x, y, and z.

Key Concepts

substitution method

The substitution method is a technique to solve systems of equations by substituting one variable with an expression involving the other variable(s). It is especially useful when one equation can easily isolate a variable. This method involves a few key steps:

• Identify an equation that you can solve for one variable in terms of the other.
• Substitute this expression into the other equations.
• Simplify and solve for the remaining variables.
• Back-substitute to find the initial variable.

For the given exercise, the substitution method starts with substituting \(u = \frac{1}{x}\), \(v = \frac{1}{y}\), and \(w = \frac{1}{z}\). Then we solve the mutated system. For instance, from the first two manipulated equations, we isolate \(u\) as \(u = -6 - 4v\). We substitute this into other equations to simplify further. Each step is followed to reduce complex systems into simpler forms which can be solved step by step.

elimination method

The elimination method involves combining equations to cancel out one of the variables. This method often simplifies solving systems with more than two equations. Here’s a detailed look at the process:

• Align like terms of all equations.
• Multiply or divide equations to match coefficients of one variable.
• Add or subtract equations to eliminate one variable, creating a simpler system.
• Repeat the process to eliminate another variable if needed.

In our given problem, we start by eliminating \(v\) from the equations. By subtracting one equation from another, a simpler equation is formed: \(u + 4v = -6\). By substituting back and eliminating systematically, the remaining variables \(u\), \(v\), and \(w\) can be isolated. Next, we repeat the method to find all variable values, eventually solving for initial variables \(x\), \(y\), and \(z\) through back-substitution.

matrix methods

Matrix methods can also be employed to solve systems of equations more efficiently, especially when dealing with larger systems. This involves representing the system in matrix form and using operations to reduce the system. The steps include:

• Write the system in the form \(AX = B\), where \(A\) is the coefficient matrix, \(X\) is the variable matrix, and \(B\) is the constant matrix.
• Use Gaussian elimination or Gauss-Jordan elimination to convert \(A\) to reduced row-echelon form.
• Perform similar operations on \(B\) to find the solutions in matrix \(X\).

For our exercise, transforming it into a matrix allows one to apply these methods systematically. We form matrices to represent the coefficients and constants: \ \begin{bmatrix} 2 & 2 & -3 \ 1 & -2 & -3 \ 7 & -2 & 9 \ \text \ Note: Incorporating matrices well can sometimes simplify complex systems, shortening the pathway to the solution.