The given system consists of equations of planes. Notice that the coefficient of y in the first equation is the additive inverse of the coefficient of y in the third equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system.
x - 2y+3z=12 & (I) 2x-y-2z=5 & (II) 2x + 2y-z=4 & (III)
We can start by adding the third equation to the first equation to eliminate the z-terms.
x-2y+3z=12 & (I) 2x-y-2z=5 & (II) 2x+2y-z=4 & (III)
Having eliminated the y-variable from the first equation, we can continue by creating additive inverse coefficients for y in the second and third equations. Then, we can add or subtract these equations to eliminate y from the second equation.
Next, we use our two equations that are only in terms of x and z to solve for the value of one of the variables. We will once again apply the Elimination Method, but this time will be similar to when using it in a system with only two variables.
The solution to the system is ( 4, -1, 2). This is the singular point at which all three planes intersect.
Now, we can check our solution by substituting the values into the system.
x-2y+3z=12 & (I) 2x-y-2z=5 & (II) 2x+2y-z=4 & (III)
