Problem 41 - A Graphical Approach to College Algebra

Question

\(\begin{aligned} 2 x-y+2 z &=6 \\ -x+y+z &=0 \\ -x-3 z &=-6 \end{aligned}\)

Step-by-Step Solution

Answer

The solution is \(x = 6\), \(y = 6\), \(z = 0\).

Step 1: Solve for x from equation 3

Start by considering the third equation \(-x - 3z = -6\). To solve for \(x\), rearrange the terms to get \(x = 3z + 6\). We will use this expression for \(x\) in the other equations.

Step 2: Substitute x in equations 1 and 2

Substitute \(x = 3z + 6\) into the first equation: \(2x - y + 2z = 6\), which becomes \(2(3z + 6) - y + 2z = 6\) or \(6z + 12 - y + 2z = 6\). Simplify it to \(8z - y = -6\) or \(y = 8z + 6\).Next, substitute \(x = 3z + 6\) into the second equation: \(-x + y + z = 0\), which becomes \(-(3z + 6) + y + z = 0\) or \(-3z - 6 + y + z = 0\). Simplify it to \(y - 2z = 6\) or \(y = 2z + 6\).

Step 3: Equate expressions for y

We have two expressions for \(y\): \(y = 8z + 6\) and \(y = 2z + 6\). Equate these two expressions: \(8z + 6 = 2z + 6\). Solve for \(z\) by subtracting \(2z\) from both sides to get \(6z = 0\), giving \(z = 0\).

Step 4: Substitute back to find y and x

Use \(z = 0\) in \(y = 2z + 6\) to find \(y\). Substituting \(z = 0\) gives \(y = 6\).Next, use \(z = 0\) in \(x = 3z + 6\) to find \(x\). Substituting \(z = 0\) gives \(x = 6\).

Step 5: Verify the solution

Verify the solution by substituting \(x = 6\), \(y = 6\), and \(z = 0\) back into the original equations.1. For \(2x - y + 2z = 6\): Substitute to get \(2(6) - 6 + 2(0) = 12 - 6 = 6\), which matches.2. For \(-x + y + z = 0\): Substitute to get \(-6 + 6 + 0 = 0\), which matches.3. For \(-x - 3z = -6\): Substitute to get \(-6 - 3(0) = -6\), which matches.All equations are satisfied, confirming the solution is correct.

Key Concepts

Linear Equations

Linear equations are mathematical expressions that create a straight line when graphed on a coordinate plane. They consist of variables and constants with linear relationships. In this problem, we have three linear equations:

• \(-x - 3z = -6\)
  
• \(-x + y + z = 0\)
  
• \(2x - y + 2z = 6\)
  

These equations combine variables \(x\), \(y\), and \(z\) with constant coefficients. The goal is to find values that satisfy each equation simultaneously. Linear equations are foundational in algebra, forming the basis for modeling and solving real-world problems.

Substitution Method

The substitution method is a technique used to solve systems of equations, particularly effective for systems like linear ones. Here’s how it is applied in our problem:Start by solving one of the equations for a specific variable. For instance, from the third equation, solve for \(x\):

• Equation: \(-x - 3z = -6\)
  
• Rearrange to find \(x\): \(x = 3z + 6\)
  

Next, substitute this expression for \(x\) into the remaining equations:

• Substitute \(x = 3z + 6\) into the first equation: \(2(3z + 6) - y + 2z = 6\)
  
• This simplifies to: \(8z - y = -6\) or \(y = 8z + 6\)
  
• Also, substitute \(x = 3z + 6\) into the second equation: \(-3z - 6 + y + z = 0\)
  
• Simplifying it yields: \(y = 2z + 6\)
  

By equating these two expressions for \(y\), we can solve for \(z\). Using substitution helps simplify the problem by reducing the number of variables step-by-step.

Verification of Solutions

Verifying solutions is the final step in solving systems of equations. It confirms that the solutions satisfy all given equations.For this problem, once we found \(x = 6\), \(y = 6\), and \(z = 0\), we need to substitute these values back into the original equations to verify:

• Substitute into \(2x - y + 2z = 6\):
  \(2(6) - 6 + 2(0) = 12 - 6 = 6\)
  This equation is satisfied.
  
• Check \(-x + y + z = 0\):
  \(-6 + 6 + 0 = 0\)
  This equation checks out as well.
  
• Confirm with \(-x - 3z = -6\):
  \(-6 - 3(0) = -6\)
  Matched exactly.
  

If all equations are true with the found values, it verifies the correctness of the solution. This step is essential to ensure there are no errors in the calculations and that the solutions are accurate and reliable.