Problem 17 - Intermediate Algebra

Question

Solve each system. $$ \left\\{\begin{array}{l} 3 x+2 y-5 z=3 \\ 4 x-2 y-3 z=-10 \\ 5 x-2 y-2 z=-11 \end{array}\right. $$

Step-by-Step Solution

Answer

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

Step 1: Analyze the System of Equations

The given system consists of three equations with three variables (x, y, z). Our goal is to find the values of x, y, and z that satisfy all three equations simultaneously.

Step 2: Eliminate Variable y

To eliminate the variable y, let's add the first equation to the second equation.Equation 1: \(3x + 2y - 5z = 3\)Equation 2: \(4x - 2y - 3z = -10\)Adding these equations: \[(3x + 2y - 5z) + (4x - 2y - 3z) = 3 - 10\] This simplifies to: \[7x - 8z = -7 \] We now have one equation without y: Equation 4: \(7x - 8z = -7\).

Step 3: Eliminate y Again

Next, let's add the first and third equations to further eliminate y.Equation 1: \(3x + 2y - 5z = 3\)Equation 3: \(5x - 2y - 2z = -11\)Adding these equations: \[(3x + 2y - 5z) + (5x - 2y - 2z) = 3 - 11\] This simplifies to: \[8x - 7z = -8\]We now have a second equation without y: Equation 5: \(8x - 7z = -8\).

Step 4: Solve for x and z

Now, let's solve Equations 4 and 5 to find x and z.Equation 4: \(7x - 8z = -7\)Equation 5: \(8x - 7z = -8\)We can use the elimination method again to solve for x and z. Multiply Equation 4 by 8 and Equation 5 by 7 to eliminate z.\(8(7x - 8z) = 8(-7)\)\(7(8x - 7z) = 7(-8)\)This gives:\[56x - 64z = -56\]\[56x - 49z = -56\]Subtract these results:\[(-64z) - (-49z) = 0\] \[-15z = 0\] Hence, \(z = 0\). Substitute \(z = 0\) back into Equation 4 or 5. Using Equation 4:\[7x - 8(0) = -7\] \[7x = -7\] \(x = -1\).

Step 5: Solve for y

Now that we have \(x = -1\) and \(z = 0\), we can substitute these values back into one of the original equations to solve for y. We'll use Equation 1:\(3x + 2y - 5z = 3\)Substitute \(x = -1\) and \(z = 0\):\[3(-1) + 2y - 5(0) = 3\]\[-3 + 2y = 3\] Adding 3 to both sides gives:\[2y = 6\] \(y = 3\).

Step 6: Verify the Solution

Finally, let's verify the solution \(x = -1\), \(y = 3\), \(z = 0\) in all three original equations:Equation 1: \(3(-1) + 2(3) - 5(0) = 3\), which simplifies to 3 = 3 (True).Equation 2: \(4(-1) - 2(3) - 3(0) = -10\), which simplifies to -10 = -10 (True).Equation 3: \(5(-1) - 2(3) - 2(0) = -11\), which simplifies to -11 = -11 (True).Since all equations are satisfied, our solution is verified.

Key Concepts

Elimination Method

The elimination method is a strategic approach for solving systems of equations. It's often used when you have multiple equations with several variables, such as in our system with the variables \(x\), \(y\), and \(z\). The primary goal is to simplify the system by eliminating one of the variables, thus reducing the number of variables in the equations. This is achieved by combining equations to cancel out a specific variable.

A significant advantage of the elimination method is its ability to handle situations where substitution may be too complex or cumbersome, especially when dealing with systems that have more than two variables. In our case, we initially focused on eliminating the variable \(y\). By adding or subtracting equations, we successfully created two new equations that only contained \(x\) and \(z\). These simplified equations are crucial because they allow us to tackle the problem with fewer variables.

Remember, while eliminating a variable, ensure that you perform identical operations on both sides of the equations. This maintains the balance of the equation, which is vital for reaching the correct solution.

Variables x, y, z

When tackling a system of equations, each variable represents an unknown term we need to solve for. In our system, we have three variables: \(x\), \(y\), and \(z\). Understanding what these variables represent is essential.

Each equation in the system provides a different perspective on how these variables are interrelated. Our job is to find a set of values for \(x\), \(y\), and \(z\) that satisfy all equations simultaneously. It often involves manipulating the equations to eliminate a variable, making it easier to solve for the others.

Consider these pointers when handling variables in systems of equations:

• Identify which variable is easiest to eliminate based on coefficients.
• Use one change per step to prevent mistakes.
• Always check your arithmetic accuracy, as small errors can lead to incorrect solutions.

By methodically eliminating variables, our problem becomes increasingly simpler to solve, leading us to the values that solve the entire system.

Solution Verification

Verifying the solution is the final and crucial step in solving systems of equations. It's the process where we substitute the determined values of the variables back into the original equations to ensure all equations are satisfied.

In our exercise, after solving for \(x = -1\), \(y = 3\), and \(z = 0\), we substituted these values back into the original three equations. Verification involves calculating both sides of each equation to confirm they match. If they do match, it confirms our solutions are correct.

During verification, remember to:

• Carefully substitute each variable's value into each equation.
• Re-calculate each equation independently to spot errors that might have been overlooked.
• Ensure that the left-hand side of the equation equals the right-hand side consistently for all equations.

This process not merely confirms your solutions, but also enhances confidence in your answer. It's a safety net ensuring no calculation errors impede the final answer. Trust the method: verification always rewards methodical and accurate solving of the equations.