Solve the system of equations by using row operations to write a matrix in REDUCED row-echelon form.
$4x+2y+3z=6$
$x+y+z=3$
$5x+y+2z=5$

• A. $x=0,y=-3,z=4$
• B. $x=-\frac{1}{4},y=-\frac{19}{4},z=\frac{11}{2}$
• C. $x=1,y=4,z=-2$
• D. $x=\frac{1}{2},y=\frac{1}{2},z=1$

Answer: (C)

Verified step by step guidance

1. Step 1: Write the system of equations as an augmented matrix. The system is: 4x + 2y + 3z = 6, x + y + z = 3, 5x + y + 2z = 5. The augmented matrix is: [[4, 2, 3, 6], [1, 1, 1, 3], [5, 1, 2, 5]].
2. Step 2: Use row operations to get a leading 1 in the first row, first column. You can do this by swapping the first row with the second row, resulting in: [[1, 1, 1, 3], [4, 2, 3, 6], [5, 1, 2, 5]].
3. Step 3: Eliminate the first column below the leading 1 by using row operations. Subtract 4 times the first row from the second row, and 5 times the first row from the third row. This gives: [[1, 1, 1, 3], [0, -2, -1, -6], [0, -4, -3, -10]].
4. Step 4: Get a leading 1 in the second row, second column by dividing the second row by -2. The matrix becomes: [[1, 1, 1, 3], [0, 1, 0.5, 3], [0, -4, -3, -10]].
5. Step 5: Eliminate the second column below the leading 1 in the second row by adding 4 times the second row to the third row. This results in: [[1, 1, 1, 3], [0, 1, 0.5, 3], [0, 0, -1, 2]]. Now, back substitute to find the values of x, y, and z.