Exercise 36 Page 172

The given system consists of equations of planes. Let's use the Elimination Method to find a solution to this system. Notice that in the third equation there is no c-term. 4a+2b+c=2 & (I) 5a-3b+2c=17 & (II) a-5b=3 & (III) Currently, none of the terms in this system will cancel out. However, if we multiply (I) by -2 the coefficient of c in this equation will be the additive inverse of the coefficient of c in the second equation; they will add to be 0. -2(4a+2b+c)=-2(2) 5a-3b+2c=17 a-5b=3 ⇓ -8a-4b - 2c=-4 5a-3b + 2c=17 a-5b=3 We can start by adding the second equation to the first equation to eliminate the c-terms.

Next, we will use our two equations that are only in terms of a and b 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 b=-1, we can substitute it into the first equation to find the value of a.

The value of a is -2. Let's substitute the value of a and b into the second equation to find c.

The solution to the system is (-2,-1,12). This is the singular point at which all three planes intersect.