Question

determine whether the two systems of linear equations yield the same solution. if so, find the solution using manone.)
(a)
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$$\begin{cases} x + 3y - z = 20 \\\\ y + 6z = -24 \\\\ z = -5 \\end{cases}$$

(b)
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$$\begin{cases} x - y + 3z = -24 \\\\ y - 2z = 16 \\\\ z = -5 \\end{cases}$$

$(x, y, z) = (\square)$

Explanation:

Step1: Solve System (a)

We start with the third equation \( z = -5 \). Substitute \( z = -5 \) into the second equation \( y + 6z = -24 \):
\( y + 6(-5) = -24 \)
\( y - 30 = -24 \)
\( y = -24 + 30 = 6 \)

Now substitute \( y = 6 \) and \( z = -5 \) into the first equation \( x + 3y - z = 20 \):
\( x + 3(6) - (-5) = 20 \)
\( x + 18 + 5 = 20 \)
\( x + 23 = 20 \)
\( x = 20 - 23 = -3 \)

So the solution for system (a) is \( x = -3 \), \( y = 6 \), \( z = -5 \).

Step2: Solve System (b)

We start with the third equation \( z = -5 \). Substitute \( z = -5 \) into the second equation \( y - 2z = 16 \):
\( y - 2(-5) = 16 \)
\( y + 10 = 16 \)
\( y = 16 - 10 = 6 \)

Now substitute \( y = 6 \) and \( z = -5 \) into the first equation \( x - y + 3z = -24 \):
\( x - 6 + 3(-5) = -24 \)
\( x - 6 - 15 = -24 \)
\( x - 21 = -24 \)
\( x = -24 + 21 = -3 \)

So both systems have the same solution.

Answer:

\((-3, 6, -5)\)