The given system consists of equations of planes. Notice that the coefficient of z in the first equation is the additive inverse of the coefficient of z in the second equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system.
-2x+y - z=2 & (I) - x-3y + z=-10 & (II) 3x+6z=-24 & (III)
We can start by adding the second equation to the first equation to eliminate the z-terms.
-2x+y-z=2 & (I) - x-3y+z=-10 & (II) 3x+6z=-24 & (III)
Having eliminated the z-variable from the first equation, we can continue by creating additive inverse coefficients for z in the second and third equations. Then, we can add or subtract these equations to eliminate z from the third equation.
Next, we will use our two equations that are only in terms of x and y to solve for the value of one of the variables. We will once again apply the Elimination Method, but this time it will be similar to when using it in a system with only two variables.
The solution to the system is ( 2, 1, -5). This is the singular point at which all three planes intersect.
Now, we can check our solution by substituting the values into the system.
-2x+y-z=2 & (I) - x-3y+z=-10 & (II) 3x+6z=-24 & (III)
