Algebra Know-It-ALL

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