284 Larger Linear Systems
one? Or between the first and the third, and then between the second and the third?” The answer is, “We
can! Either of those alternative schemes will work just as well as the one we chose. The intermediate equa-
tions will differ, but the ultimate solution to the three-by-three system will turn out the same.”
Whenever you want to solve a three-by-three system of equations, you must somehow involve all three
of the equations in the solution process. As you continue to study algebra and take more advanced courses,
you’ll learn other ways to solve three-by-three systems than the methods presented in this book. You might
also learn techniques to solve four-by-four or larger systems. In any case, if you want to find the solution
to a system of equations, you must give every one of those equations some “say” in the outcome.Here’s a challenge!
Put the original equations in our three-by-three linear system into a form where z appears all by itself on
the left sides of the equals signs, and expressions containing only constants, x, and y appear on the right
sides. This defines two independent variables (in this case x and y) and a single dependent variable (z).
Here are the original equations, for reference:− 4 x+ 2 y− 3 z= 5
2 x− 5 y=z− 1
3 x=− 6 y+ 7 zSolution
To morph these equations, we can use the same sort of algebra that gets two-variable linear equations into
SI form. Here are the processes, step-by-step. You have learned enough algebra so you can follow along
without detailed explanations.− 4 x+ 2 y− 3 z= 5
2 y− 3 z= 4 x+ 5
− 3 z= 4 x− 2 y+ 5
−z= (4/3)x− (2/3)y+ 5/3
z= (−4/3)x+ (2/3)y− 5/32 x− 5 y=z− 1
2 x− 5 y+ 1 =z
z= 2 x− 5 y+ 13 x=− 6 y+ 7 z
3 x+ 6 y= 7 z
(3/7)x+ (6/7)y=z
z= (3/7)x+ (6/7)yThese three equations represent relations that map pairs of variables (in this case x and y) into a single
variable (in this case z). If we call the relations f, g, and h, then we can write:f (x,y)= (−4/3)x+ (2/3)y− 5/3
g (x,y)= 2 x− 5 y+ 1
h (x,y)= (3/7)x+ (6/7)y