provide very short feedback without "your" to this "Let's start by creating three equations in the form ax by cz=d, where a,b,c, and d are constants between -5 and 5: 1. 2x 3yz=4 2. 4xy 2z=2 3. x 2y z=5 Now, let's perform row operations to bring the system to row-echelon form and find the solution: 1. Row Operation 1 (R1 = R1/2): 1/2(2x 3yz=4) 4xy 2z=2 x 2y z=5 The system becomes: x 3/2y1/2z=2 4xy 2z=-2 x 2y z=2=5 2. Row Operation 2 (R2 = R2 - 4R1, R3 = R3 - R1): x 3/2y1/2z=2 5y 4z=-10 5/2y 3/2z=2=3 3. Row Operation 3 (R3 = R3 - \frac{5}{2}R2): x 3/2y1/2z2 -5y 4z=10 Z=8 Now, let's enter these equations into the GeoGebra 3D calculator and observe the graphs. Observations: The system represents three planes in 3D space. The first two equations create planes, and the third equation defines a line. The point of intersection of these three planes/line is the solution to the system. 1. Equation 1 (Plane): x 3/2y21z=2 2. Equation 2 (Plane): 5y 4z=10 3. Equation 3 (Line): z=8 The GeoGebra visualization shows the three planes intersecting in a line, which is consistent with our row-echelon form and