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Use​ left-to-right elimination to solve the system of equations.

left brace Start 2 By 1 Matrix 1st Row 1st Column 2nd Row 1st Column EndMatrix

x minus 4 y plus 4 z equals
27
y minus 3 z equals
negative 28
y plus 3 z equals
32
Question content area bottom
Part 1
Select the correct choice below and fill in any answer boxes within your choice.
A.
The one solution is xequals
  
enter your response here​, yequals
  
enter your response here​, and zequals
  
enter your response here. ​(Simplify your​ answers.)
B.
There are infinitely many solutions. If z is allowed to be any real​ number, then xequals
  
enter your response here and yequals
  
enter your response here.
​(Type expressions using z as the​ variable.)
C.

1 answer

To solve the given system of equations using left-to-right elimination, we start by identifying the equations:

  1. \( x - 4y + 4z = 27 \) (Equation 1)
  2. \( y - 3z = -28 \) (Equation 2)
  3. \( y + 3z = 32 \) (Equation 3)

Step 1: Solve for \( y \) from Equations 2 and 3

Let's start with Equation 2 and Equation 3.

From Equation 2: \[ y = 3z - 28 \]

From Equation 3: \[ y = 32 - 3z \]

Now we can set these two equations for \( y \) equal to one another: \[ 3z - 28 = 32 - 3z \]

Step 2: Solve for \( z \)

Combine like terms: \[ 3z + 3z = 32 + 28 \] \[ 6z = 60 \] \[ z = 10 \]

Step 3: Find \( y \)

Now substitute \( z = 10 \) back into either Equation 2 or Equation 3. We'll use Equation 2: \[ y = 3(10) - 28 \] \[ y = 30 - 28 \] \[ y = 2 \]

Step 4: Find \( x \)

Now substitute \( y = 2 \) and \( z = 10 \) back into Equation 1: \[ x - 4(2) + 4(10) = 27 \] \[ x - 8 + 40 = 27 \] \[ x + 32 = 27 \] \[ x = 27 - 32 \] \[ x = -5 \]

Conclusion

Now that we have found the values:

  • \( x = -5 \)
  • \( y = 2 \)
  • \( z = 10 \)

Final Answer

The one solution is:

  • \( x = -5 \),
  • \( y = 2 \),
  • \( z = 10 \).

Thus, the correct choice is A. Select A and fill in any answer boxes:

  • The one solution is \( x = -5 \), \( y = 2 \), and \( z = 10 \).