Questions: The figure on the right shows the intersections of three one-way streets. Use Gaussian elimination to solve the system of equations that models this situation (shown below). x+10=y+15 z+12=x+9 y+24=z+22 Select the correct choice below and fill in any answer boxes within your choice. A. There is one solution. The solution set is ). (Simplify your answers.) B. There are infinitely many solutions. The solution set is (, ), where z is any real number. (Type expressions using z as the variable.) C. There is no solution.

Solution Steps

The given system of equations is:

1. \( x + 10 = y + 15 \)
2. \( z + 12 = x + 9 \)
3. \( y + 24 = z + 22 \)

Simplify each equation to standard form:

1. \( x - y = 5 \)
2. \( z - x = -3 \)
3. \( y - z = -2 \)

Set up the augmented matrix for the system of equations: \[ \begin{pmatrix} 1 & -1 & 0 & | & 5 \\ -1 & 0 & 1 & | & -3 \\ 0 & 1 & -1 & | & -2 \\ \end{pmatrix} \]

Perform row operations to reduce the matrix to row echelon form:

1. Swap Row 1 and Row 2: \[ \begin{pmatrix} -1 & 0 & 1 & | & -3 \\ 1 & -1 & 0 & | & 5 \\ 0 & 1 & -1 & | & -2 \\ \end{pmatrix} \]
2. Multiply Row 1 by -1: \[ \begin{pmatrix} 1 & 0 & -1 & | & 3 \\ 1 & -1 & 0 & | & 5 \\ 0 & 1 & -1 & | & -2 \\ \end{pmatrix} \]
3. Subtract Row 1 from Row 2: \[ \begin{pmatrix} 1 & 0 & -1 & | & 3 \\ 0 & -1 & 1 & | & 2 \\ 0 & 1 & -1 & | & -2 \\ \end{pmatrix} \]
4. Add Row 2 to Row 3: \[ \begin{pmatrix} 1 & 0 & -1 & | & 3 \\ 0 & -1 & 1 & | & 2 \\ 0 & 0 & 0 & | & 0 \\ \end{pmatrix} \]

The last row indicates that the system has infinitely many solutions because it reduces to \(0 = 0\).

Final Answer