258 Two-by-Two Linear Systems
The terms eby and −bey are additive inverses, so they disappear from the sum. Next, we can apply the
distributive law “backward” in the left side, obtaining
(ea−bd)x=ec−bf
We can solve for x if we divide through by (ea−bd), assuming that (ea−bd)≠ 0:
x= (ec−bf )/(ea−bd)
Are you astute?
Have you noticed that the “taboos” in the above derivations are actually two different ways of saying the
same thing? They’re stated like this:
(db−ae)≠ 0
and
(ea−bd)≠ 0
Both of these inequalities are equivalent to the statement ae≠bd. Can you see why? What do you think
will happen if a linear system has coefficients in the above form such that ae=bd? You’ll get a chance to
explore a situation like that in exercises 5, 6, and 7 at the end of this chapter!
Rename and Replace
A two-by-two linear system can be unraveled by renaming one variable in terms of the other,
and then creating a single-variable, first-degree equation from the result. We solve that equa-
tion, and then plug the number into a strategic spot to solve for the other variable. This pro-
cess is usually called the substitution method. I like to call it rename and replace.
Rename one variable
When we want to solve a two-by-two linear system by rename-and-replace, we begin by mor-
phing one of the equations into SI form. Consider this pair of equations:
− 7 v+w+ 10 = 0
and
4 v+ 8 w=− 40
We can take the first equation and add 7v to each side, getting
w+ 10 = 7 v