Problem 46 - College Algebra with Modeling and Visualization

Question

Solve the system, if possible. $$ \begin{aligned} &4 x-y-z=0\\\ &4 x-2 y \quad=0\\\ &2 x \quad+z=1 \end{aligned} $$

Step-by-Step Solution

Answer

The solution is \(x = \frac{1}{4}\), \(y = \frac{1}{2}\), and \(z = \frac{1}{2}\).

Step 1: Start with the system

The given system is:\[\begin{align*}4x - y - z &= 0 \4x - 2y &= 0 \2x + z &= 1\end{align*}\]We will solve this system using substitution or elimination methods.

Step 2: Solve for a variable

Let's start with the second equation, which is simpler. Solve for one of the variables. From:\[4x - 2y = 0 \]Divide the entire equation by 2:\[2x - y = 0 \]Thus, we can express \(y\) in terms of \(x\):\[y = 2x\]

Step 3: Substitute into another equation

Substitute \(y = 2x\) into the first equation:\[4x - (2x) - z = 0\]Simplify the equation:\[2x - z = 0\]Thus, we can express \(z\) in terms of \(x\):\[z = 2x\]

Step 4: Substitute into the third equation

Using the expression \(z = 2x\), substitute into the third equation:\[2x + (2x) = 1\]Simplify the equation:\[4x = 1\]Solve for \(x\):\[x = \frac{1}{4} \]

Step 5: Find the values of other variables

Now that we have \(x = \frac{1}{4}\), substitute it back into the expressions for \(y\) and \(z\):For \(y\):\[y = 2x = 2\left(\frac{1}{4}\right) = \frac{1}{2}\]For \(z\):\[z = 2x = 2\left(\frac{1}{4}\right) = \frac{1}{2}\]

Step 6: Verify the solution

Substitute \(x = \frac{1}{4}\), \(y = \frac{1}{2}\), and \(z = \frac{1}{2}\) back into the original equations to verify:1. For \(4x - y - z = 0\): \[ 4\left(\frac{1}{4}\right) - \frac{1}{2} - \frac{1}{2} = 1 - 1 = 0 \]2. For \(4x - 2y = 0\): \[ 4\left(\frac{1}{4}\right) - 2\left(\frac{1}{2}\right) = 1 - 1 = 0 \]3. For \(2x + z = 1\): \[ 2\left(\frac{1}{4}\right) + \frac{1}{2} = \frac{1}{2} + \frac{1}{2} = 1 \]All equations are satisfied.

Key Concepts

Substitution Method

The substitution method is a popular technique to solve linear systems. It involves expressing one variable in terms of another by transforming one of the equations. This method allows us to substitute and simplify step-by-step until we find the value of each variable.

Here's how the substitution method works:

• First, choose one equation and solve for one variable in terms of another variable.
• Substitute this expression into the other equations. This will help to eliminate one variable and solve for the remaining variable(s).
• Continue substituting until all the variables have specific values.
• In our exercise, we solved for \( y \) in terms of \( x \) as \( y = 2x \) from the second equation. We used this relation to express \( z \) in terms of \( x \) and finally determined \( x = \frac{1}{4} \).

The substitution method is straightforward, especially when dealing with equations where isolating a variable is easy. This method is particularly useful when the equations are aligned in such a way that one variable can be eliminated early in the process.

Elimination Method

The elimination method, another powerful approach for solving systems of equations, focuses on canceling out one of the variables by adding or subtracting equations. This technique can sometimes simplify the solution process even more than substitution.

Here's a brief outline of how the elimination method works:

• Select two equations from the system that are easy to manipulate to make the coefficients of one variable identical.
• Use addition or subtraction to eliminate the selected variable across these equations.
• This gives you a simpler equation that you can use to solve for the remaining variable(s).
• In our problem, elimination wasn't directly used, but it could have been applied by manipulating the coefficients of \( y \) or \( z \) in two equations to get them to match or cancel out.

The elimination method works best for systems where direct cancellation of variables can be achieved without extensive fraction or decimal calculations.

Verification of Solutions

After solving a linear system using either method, it is essential to verify the solutions. Verification confirms that the proposed solution satisfies all original equations. Without verification, there's a risk of proceeding with incorrect values, so always double-check!

This is how you can verify your solutions:

• Take the values obtained for each variable and plug them back into the original equations.
• Check if each equation holds true with the solved values.
• If every original equation is satisfied, your solution is verified as correct.
• In our exercise, substituting \( x = \frac{1}{4} \), \( y = \frac{1}{2} \), and \( z = \frac{1}{2} \) back into all three given equations confirmed that the original system was satisfied.

Verification is a crucial final step when solving linear systems. It serves as a self-check to ensure that the solutions derived are correct and consistent with the problem.