Question

Solve the system of equations.
left brace Start 2 By 1 Matrix 1st Row 1st Column 2nd Row 1st Column EndMatrix

x minus y minus 8 z equals
negative 81
y plus 5 z equals
47
6 y plus 7 z equals
52
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.
There is no solution.

Answer 1

To solve the given system of equations, we will translate the equations and arrange them systematically.

The equations given are:

1. \( x - y - 8z = -81 \) (Equation 1)
2. \( y + 5z = 47 \) (Equation 2)
3. \( 6y + 7z = 52 \) (Equation 3)

Let's solve this system step-by-step:

**Step 1:** Solve Equation 2 for \( y \).

From Equation 2:

\[
y = 47 - 5z
\]

**Step 2:** Substitute \( y \) in Equations 1 and 3.

For Equation 1:

\[
x - (47 - 5z) - 8z = -81
\]

Simplifying this gives:

\[
x - 47 + 5z - 8z = -81
\]
\[
x - 47 - 3z = -81
\]
\[
x - 3z = -81 + 47
\]
\[
x - 3z = -34
\]
\[
x = -34 + 3z \quad \text{(Equation 4)}
\]

For Equation 3:

Now substitute \( y = 47 - 5z \) into Equation 3:

\[
6(47 - 5z) + 7z = 52
\]
\[
282 - 30z + 7z = 52
\]
\[
282 - 23z = 52
\]
\[
-23z = 52 - 282
\]
\[
-23z = -230
\]
\[
z = 10
\]

**Step 3:** Substitute \( z \) back to find \( y \) and \( x \).

Now substitute \( z = 10 \) back into Equation 2 to find \( y \):

\[
y = 47 - 5(10) = 47 - 50 = -3
\]

Now substitute \( y \) and \( z \) into Equation 4 to find \( x \):

\[
x = -34 + 3(10) = -34 + 30 = -4
\]

**Final Solution:**

We find:

\[
x = -4, \quad y = -3, \quad z = 10
\]

Thus, the one solution is:

A. The one solution is \( x = -4 \), \( y = -3 \), and \( z = 10 \).