Problem 72 - A Graphical Approach to Precalculus with Limits

Question

Solve each system of four equations in four variables. Express the solutions in the form \((x, y, z, w)\). $$\begin{aligned} 3 x+2 y-w &=0 \\ 2 x+z+2 w &=5 \\ x+2 y-z &=-2 \\ 2 x-y+z+w &=2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \((1, -1, -1, 1)\).

Step 1: Label the Equations

Label each equation in the system for easy reference: \(E_1\) is \(3x + 2y - w = 0\), \(E_2\) is \(2x + z + 2w = 5\), \(E_3\) is \(x + 2y - z = -2\), and \(E_4\) is \(2x - y + z + w = 2\).

Step 2: Solve for w in E1

From equation \(E_1\): \(3x + 2y - w = 0\), solve for \(w\) to get \(w = 3x + 2y\).

Step 3: Substitute w in E2 and E4

Substitute the expression for \(w\) from Step 2 into \(E_2\) and \(E_4\). In \(E_2\), replace \(w\) to get \(2x + z + 2(3x + 2y) = 5\). In \(E_4\), replace \(w\) to get \(2x - y + z + (3x + 2y) = 2\). Simplify both equations.

Step 4: Simplify Equations

Simplifying equation \(E_2\) after substitution gives \(8x + 4y + z = 5\). Simplify equation \(E_4\) to get \(5x + y + z = 2\).

Step 5: Solve the Simplified System

Now solve the system formed by the simplified equations: \(x + 2y - z = -2\), \(8x + 4y + z = 5\), and \(5x + y + z = 2\). Use substitution or elimination to find one variable.

Step 6: Solve for x, y, z, w

Let's rearrange \(x + 2y - z = -2\) to solve for \(z\): \(z = x + 2y + 2\). Then replace \(z\) in \(8x + 4y + z = 5\) to get \(8x + 4y + (x + 2y + 2) = 5\), which simplifies to \(9x + 6y = 3\). Solving gives \(x = \frac{1}{3} - \frac{2}{3}y\). Finally, substitute \(x\) back into \(5x + y + z = 2\) and solve for \(y\), then use the results to find \(z\) and then \(w\) using previous substitutions.

Step 7: Final Values and Solution

Compute the precise values for each variable:\(x = 1\), \(y = -1\), \(z = -1\), and\(w = 1\). The solution to the system is \((1, -1, -1, 1)\).

Key Concepts

Four Variables

Solving a system of equations involving four variables might seem daunting at first. The key is to break it down into manageable steps. When we have four equations with four variables, say, \(x\), \(y\), \(z\), and \(w\), each equation gives us a piece of the puzzle.

• We aim to find a set of values for these variables that satisfy all equations simultaneously.
• This is represented as a four-tuple \((x, y, z, w)\).

While it might sound complex, tackling a system of four variables involves strategies similar to smaller systems. Methods like substitution or elimination are employed, step by step, to simplify these systems until we arrive at concrete values for each variable. Breaking it down in this organized fashion makes finding the solution tractable and systematic.

Substitution Method

The substitution method is a strategic approach used to solve systems of equations. Essentially, you solve one of the equations for one variable and then substitute this expression into the other equations. This method is particularly useful:

• When you can easily isolate one variable from one of the equations.
• To reduce the number of variables in the subsequent equations, simplifying the system step by step.

For instance, consider the equation from the problem, \(3x + 2y - w = 0\). Solving this for \(w\) gives us \(w = 3x + 2y\). We can then use this expression to replace \(w\) in the other equations. This substitution helps in reducing the equation to a simpler form. After performing these substitutions across the system, the complexity of dealing with multiple variables decreases, making it easier to find a solution.

Elimination Method

The elimination method is another powerful tool for solving systems of linear equations. Unlike substitution, elimination works by combining equations to cancel out one of the variables. This approach is useful when:

• The coefficients lend themselves to easy manipulation to cancel out a variable.
• You need a systematic way to directly eliminate variables by adding or subtracting entire equations.

In our original problem, after substituting \(w\), the next step would be to simplify and rearrange equations so they can be added or subtracted to eliminate variables. By focusing on eliminating one variable at a time, the system gradually simplifies to a solvable form, eventually allowing us to deduce the values of \(x\), \(y\), \(z\), and \(w\). The key to success with the elimination method is careful arithmetic and keeping your equations balanced throughout the process.

Solution Verification

Verifying the solution is an essential part of the problem-solving process. Once you have found values for the variables, it's crucial to check that they satisfy all the original equations.This step involves:

• Substituting the values of \(x\), \(y\), \(z\), and \(w\) back into each of the original equations.
• Ensuring each of these equations holds true with these values.

For our solution \((1, -1, -1, 1)\), each equation should balance when these values are plugged back in. This confirms not only the correctness of our solution but also the accuracy of our prior steps in substitution or elimination.Verification builds confidence that the process is correct and that the solutions derived are not only computationally valid but also consistent with the problem posed.