Problem 50 - Elementary and Intermediate Algebra

Question

Solve. $$\begin{aligned} &w+x+y+z=2,\\\ &w+2 x+2 y+4 z=1,\\\ &w-x+y+z=6,\\\ &w-3 x-y+z=2 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \(w = 3\), \(x = -2\), \(y = 1\), and \(z = 0\).

Step 1 - Write the System of Equations

Consider the given system of linear equations: 1. \(w + x + y + z = 2\)2. \(w + 2x + 2y + 4z = 1\)3. \(w - x + y + z = 6\)4. \(w - 3x - y + z = 2\). We'll solve this system by using methods such as substitution or elimination.

Step 2 - Express all Equations in Terms of w

To simplify, let's first express each of the equations in terms of w: 1. \(w = 2 - x - y - z\)2. \(w = 1 - 2x - 2y - 4z\)3. \(w = 6 + x - y - z\)4. \(w = 2 + 3x + y - z\). Now we can set these expressions for \(w\) equal to each other and solve for the other variables.

Step 3 - Equate the Expressions

Set the expressions for \(w\) from Step 2 equal to each other to create new equations. Equate (1) and (2): \(2 - x - y - z = 1 - 2x - 2y - 4z\).Simplify to get: \(1 + x + y + 3z = 0\) or \(x + y + 3z = -1\). This is our first new equation.

Step 4 - Simplify by Equating Other Expressions

Now equate expressions (2) and (3): \(1 - 2x - 2y - 4z = 6 + x - y - z\).Simplify to get: \(-5 - 3x - y - 3z = 0\) or \(3x + y + 3z = -5\). This is our second new equation.

Step 5 - Solve the System of New Equations

We now have the new system: \(x + y + 3z = -1\) and \(3x + y + 3z = -5\).Subtract the first equation from the second: \((3x + y + 3z) - (x + y + 3z) = -5 - (-1)\), which simplifies to: \(2x = -4\) or \(x = -2\).

Step 6 - Substitute Back to Find y and z

Substitute \(x = -2\) back into \(x + y + 3z = -1\): \(-2 + y + 3z = -1\), which simplifies to: \(y + 3z = 1\). Choose a value for \(z\) to solve for \(y\). For simplicity, let \(z = 0\). Thus, \(y = 1\).

Step 7 - Find w Using One of the Original Equations

Using the original equation \(w = 2 - x - y - z\): Substitute \(x = -2\), \(y = 1\), and \(z = 0\): \(w = 2 - (-2) - 1 - 0 = 2 + 2 - 1 = 3\).

Step 8 - Verify the Solution

Finally, verify the values \(w = 3\), \(x = -2\), \(y = 1\), and \(z = 0\) in all original equations to ensure all equations hold true.

Key Concepts

Substitution Method

The substitution method is a popular technique for solving systems of linear equations. It involves solving one equation for one variable and then substituting this expression into the other equations.
This reduces the number of equations and variables step by step until you're left with a single variable.

For instance, given our system of equations:

• \( w + x + y + z = 2 \)
• \( w + 2x + 2y + 4z = 1 \)
• \( w - x + y + z = 6 \)
• \( w - 3x - y + z = 2 \)

The first step in the substitution method could be to solve one of the equations for \(w\).
For example from the first equation: \
\( w = 2 - x - y - z \).
Substituting this expression for \(w\) in the remaining equations simplifies them, allowing you to isolate other variables step by step.

Elimination Method

The elimination method involves combining the equations in a system of linear equations to eliminate one of the variables.
This simplifies the system, making it easier to solve. You add or subtract equations to cancel out a variable systematically.

In our example system:

• \( w + x + y + z = 2 \)
• \( w + 2x + 2y + 4z = 1 \)
• \( w - x + y + z = 6 \)
• \( w - 3x - y + z = 2 \)

To eliminate \(w\), you can set expressions equal to each other as shown in the solution steps.
For example, combining the first and second equations: \
\( 2 - x - y - z = 1 - 2x - 2y - 4z \)
Simplifying this results in another equation with fewer variables: \
\( 1 + x + y + 3z = 0 \) or \( x + y + 3z = -1 \).
By systematically eliminating variables, you'll ultimately find the values of \(x, y, z\) and \(w\).

Linear Equations

Linear equations are algebraic expressions that represent a straight line on a graph.
They can have one or more variables, and the solutions of these equations represent the points where their graphs intersect.

A linear equation in standard form looks like: \(Ax + By + Cz + ... = D\) where \(A, B, C\) are coefficients and \(D\) is a constant.

• The given problem consists of four linear equations with four variables.
• Solving such systems involves either the substitution or elimination method to find a common solution.

For our specific system:

• Each equation represents a plane in a four-dimensional space.
• The solution to the system is the point where all planes intersect.

Understanding these fundamental concepts will help you apply appropriate methods to solve linear equations efficiently and accurately.