Problem 24 - A Graphical Approach to Precalculus with Limits

Question

Solve each system analytically. If the equations are dependent, write the solution set in terms of the variable \(z\). $$\begin{aligned} x+y+z &=3 \\ 3 x-3 y-4 z &=-1 \\ x+y+3 z &=11 \end{aligned}$$

Step-by-Step Solution

Answer

The solution is \( x = 2 \), \( y = -3 \), and \( z = 4 \).

Step 1: Label the Equations

Let's label the given system of equations for easy reference.\( (1) \) \( x + y + z = 3 \)ewline\( (2) \) \( 3x - 3y - 4z = -1 \)ewline\( (3) \) \( x + y + 3z = 11 \)

Step 2: Eliminate y in Equation (2) and (3)

Subtract equation (1) from equation (3) to eliminate \(y\):ewline\( (x + y + 3z) - (x + y + z) = 11 - 3 \)ewline Simplifying gives \( 2z = 8 \), so \( z = 4 \).

Step 3: Substitute z into Equations

Substitute \( z = 4 \) into equations (1) and (2). In equation (1):ewline\( x + y + 4 = 3 \) gives \( x + y = -1 \).ewline In equation (2): \( 3x - 3y - 4(4) = -1 \) gives \( 3x - 3y - 16 = -1 \), simplifying to \( 3x - 3y = 15 \).

Step 4: Simplify and Solve for x and y

From \( 3x - 3y = 15 \), we can simplify to \( x - y = 5 \). Now we have two equations:ewline\( x + y = -1 \)ewline\( x - y = 5 \).ewline Add these two equations to solve for \(x\):ewline \( 2x = 4 \), hence \( x = 2 \).

Step 5: Solve for y

Substitute \( x = 2 \) back into \( x + y = -1 \):ewline\( 2 + y = -1 \). Therefore, \( y = -3 \).

Step 6: Confirm the Solution

We now have \( x = 2 \), \( y = -3 \), and \( z = 4 \). Substitute these values back into the original equations to verify correctness: ewline\[\begin{aligned} x+y+z &=2-3+4=3 \,\text{(True)} \ 3x-3y-4z &=3(2) -3(-3)-4(4)=6+9-16=-1 \,\text{(True)} \ x+y+3z &=2-3+12=11 \,\text{(True)} \end{aligned}\] All check out, so the solution is consistent.

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that deals with vectors, vector spaces, and systems of linear equations. Understanding this topic is essential because many real-world problems can be expressed as a system of equations, just like in our Original Exercise. By handling a system of equations, one can find values for variables that satisfy all the given equations simultaneously. In linear algebra:

• Variables are often represented as vectors.
• Equations can be described by matrices, enhancing computation efficiency.
• Solutions can show intersections where all equations hold true.

In the exercise, we are dealing with a 3-variable system. Each equation represents a plane in three-dimensional space. The solution gives us the point where all three planes intersect. By executing systematic steps, like labeling the equations and simplifying terms, the system becomes more manageable.

Substitution Method

The substitution method is a popular technique for solving systems of equations, especially when one of the equations is easily solvable for one variable. This method involves isolating one variable in one of the equations and then substituting that expression into the other equations. This way, we 'substitute' the unknown variable with known values, gradually breaking down the system.

• Start by solving one equation for a variable.
• Replace that variable in the other equations.
• Continue the process until all variables are determined.

In the provided problem, after isolating and solving for the variable \(z\) through comparison and subtraction, we substitute \(z\) into other equations. This simplifies the remaining system to two variables, which is easier to manage. This approach is systematic and step-by-step, making it a neat and organized way to unravel the solution.

Solution Verification

Once a solution is found, it is crucial to verify its correctness to ensure all work was done properly. Solution verification involves plugging the values back into the original equations to see if they hold true.

• Substitute the solution back into each equation.
• Ensure each equation balances after substitution.
• If any equation does not hold, reassess the working steps.

In the original step-by-step solution, after finding \(x = 2\), \(y = -3\), and \(z = 4\), these values were substituted back into all three equations. Each one was confirmed to be accurate, providing confidence that the solution is consistent and correct. This last step cannot be skipped, as it ensures precision and accuracy in mathematical problem-solving.