Problem 6 - Algebra 2

Question

Solve each system of equations. \(x+y+z=12\) \(6 x-2 y-z=16\) \(3 x+4 y+2 z=28\)

Step-by-Step Solution

Answer

The solution is \(x = 2\), \(y = 2\), \(z = 8\).

Step 1: Write the System of Equations

First, we need to clearly list the system of equations we are solving: 1) \(x + y + z = 12\) 2) \(6x - 2y - z = 16\) 3) \(3x + 4y + 2z = 28\).

Step 2: Eliminate One Variable

We will eliminate variable \(z\) by manipulating the equations. Let's eliminate \(z\) from equations (1) and (2).First, multiply equation (1) by 1 to keep it the same, then add to equation (2):\[(x + y + z) + (6x - 2y - z) = 12 + 16\]\[7x - y = 28 \ ext{(Equation 4)}\]

Step 3: Eliminate the Same Variable Again

Next, eliminate \(z\) using equations (1) and (3). First, multiply equation (1) by 2 to match the coefficients of \(z\):\[ 2(x + y + z) = 2(12) \] \[ 2x + 2y + 2z = 24 \] Subtract this from equation (3):\[(3x + 4y + 2z) - (2x + 2y + 2z) = 28 - 24 \]\[x + 2y = 4 \ ext{(Equation 5)}\]

Step 4: Solve the New System

Now solve the new system of equations formed by equations (4) and (5):1) \(7x - y = 28\) 2) \(x + 2y = 4\).Substitute \(y = 2\) from substituting \(x\) from equation (5) into equation (4):Using equation (5): \(y = 4 - x\).

Step 5: Substitute Back

Substitute \(x = 2\) into \(x + 2y = 4\):\[2 + 2y = 4\]\[\,\, 2y = 4 - 2\]\[y = 2\]Now substitute back to find \(z\) using equation (1):\[ 2 + 2 + z = 12 \]\[ z = 12 - 4 \]\[ z = 8 \]

Step 6: Verify the Solution

Verify the solution by substituting \(x = 2\), \(y = 2\), and \(z = 8\) back into the original system:1) \(2 + 2 + 8 = 12\)2) \(6(2) - 2(2) - 8 = 16\)3) \(3(2) + 4(2) + 2(8) = 28\)All checks are true, confirming the solution.

Key Concepts

Algebraic Solution

When tackling a system of equations, we often use algebraic methods to find the values of unknowns. Here, the goal is to solve for the variables \(x\), \(y\), and \(z\) in a set of three equations:

• \(x + y + z = 12\)
• \(6x - 2y - z = 16\)
• \(3x + 4y + 2z = 28\)

An algebraic solution involves expressing one variable in terms of others and substituting repeatedly until all variables are isolated. This example guides you through this process step-by-step. It's systematic, ensuring each calculation is verified. This technique is powerful as it helps dissect complex relationships between terms without requiring graphical aids. Whether in school or professional fields, understanding this approach enhances problem-solving skills in algebra.

Variable Elimination

Variable elimination is a key step in solving systems of equations. This is where you aim to simplify the equations by removing one variable at a time, making the system easier to solve. Let's focus on eliminating \(z\) first. We take two equations and manipulate them to cancel out \(z\), leading to a simpler system. This involves:

• Adjusting coefficients of \(z\) in the equations.
• Adding/subtracting equations.
• Seeing how these changes create a new equation where one variable is removed.

The example shows how equation combination and rearrangement yield equations like \(7x-y = 28\) and \(x + 2y = 4\). By working with fewer variables, you reduce complexity and narrow the focus to finding the precise values of \(x\) and \(y\). Consider it like peeling away layers to get to the core of the problem.

Mathematical Reasoning

Mathematical reasoning forms the backbone of solving equations through a structured method. It applies logic and careful calculation to arrive at a solution, always ensuring accuracy. In the example, we reason through steps such as:

• Checking solutions by substituting back into original equations to confirm they hold true.
• Recognizing patterns or relationships that simplify the problem.

Step-by-step, each part of the process uses reasoning to guide the solutions. When you solve for \(x = 2\), \(y = 2\), and \(z = 8\), verification ensures no errors were made. Programming your brain to verify by substitution is like using a GPS after getting directions—you reassure yourself you're on the correct path. This practice underpins not just solving equations but all math, reinforcing that every answer is a product of correct logic and calculations.