The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system.
6q-r+2s=8 & (I) 2q+3r-s=-9 & (II) 4q+2r+5s=1 & (III)
Currently, none of the terms in this system will cancel out. However, if we multiply (I) by 2 the coefficient of r in this equation will be the additive inverse of the coefficient of r in the third equation; they will add to be 0.
2(6q-r+2s)=2(8) 2q+3r-s=-9 4q+2r+5s=1
⇓
12q - 2r+4s=16 2q+3r-s=-9 4q + 2r+5s=1
We can start by adding the first equation to the third equation to eliminate the r-terms.
12q-2r+4s=16 & (I) 2q+3r-s=-9 & (II) 4q+2r+5s=1 & (III)
Having eliminated the r-variable from the third equation, we can continue by creating additive inverse coefficients for r in the first and second equations. Then, we can add or subtract these equations to eliminate r from the second equation.
Next, we will use our two equations that are only in terms of q and s 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 ( 12, -3, 1). This is the singular point at which all three planes intersect. Now, we can check our solution by substituting the values into the system.
12q-2r+4s=16 & (I) 2q+3r-s=-9 & (II) 4q+2r+5s=1 & (III)
