Exercise 9 Page 171

The given system consists of equations of planes. Notice that the coefficients of y in the first and the third equations are the additive inverse of the coefficient of y in the second equation; they will add to be 0. Let's use the Elimination Method to find a solution to this system. x - y+z=-1 & (I) x + y+3z=-3 & (II) 2x - y+2z=0 & (III) We can start by adding the second equation to the first and the third equations to eliminate the y-terms. Next, we will 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 it will be similar to when using it in a system with only two variables. Now that we know that z=-3, we can substitute it into the first equation to find the value of x. The value of x is 4. Let's substitute both values into the second equation to find y. The solution to the system is ( 4, 2, -3). This is the singular point at which all three planes intersect. Now we can check our solution by substituting the values into the system. Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.