Question

\

$$\begin{cases} -x + y + 2z = -3 \\\\ 2x - y + z = 5 \\\\ 5x - 2y + 5z = 12 \\end{cases}$$

choose the best description of its solution.
if applicable, give its solution.

\bigcirc the system has no solution.

\bigcirc the system has a unique solution.

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

the system has infinitely many solutions.

\quad (x, y, z) =

\quad\quad \bigcirc (x, \square, \square)

\quad\quad \bigcirc (\square, y, \square)

\quad\quad \bigcirc (\square, \square, z)

Explanation:

Step1: Add first two equations

Add \(-x + y + 2z = -3\) and \(2x - y + z = 5\).
\((-x + 2x)+(y - y)+(2z + z)=-3 + 5\)
\(x + 3z = 2\) (Equation 4)

Step2: Multiply first equation by 2 and add to third

Multiply \(-x + y + 2z = -3\) by 2: \(-2x + 2y + 4z = -6\).
Add to \(5x - 2y + 5z = 12\):
\((-2x + 5x)+(2y - 2y)+(4z + 5z)=-6 + 12\)
\(3x + 9z = 6\) (Equation 5)

Step3: Simplify Equation 5

Divide Equation 5 by 3: \(x + 3z = 2\), which is same as Equation 4.
So two equations are dependent, system has infinitely many solutions. Let \(z = t\) (parameter). From Equation 4: \(x = 2 - 3t\).

Substitute \(x = 2 - 3t\) and \(z = t\) into first equation:
\(-(2 - 3t)+ y + 2t = -3\)
\(-2 + 3t + y + 2t = -3\)
\(y + 5t = -1\)
\(y = -1 - 5t\)

So solutions are \((x, y, z)=(2 - 3t, -1 - 5t, t)\), or in the given form (with \(z\) as parameter, match the selected option \((\square, \square, z)\)): let \(z = z\), then \(x = 2 - 3z\), \(y = -1 - 5z\). So \((x, y, z)=(2 - 3z, -1 - 5z, z)\).

Answer:

The system has infinitely many solutions.
\((x, y, z)=(2 - 3z, -1 - 5z, z)\) (or with parameter \(z\) as given in the option format, the filled boxes would be \(2 - 3z\), \(-1 - 5z\), and \(z\) as the parameter).