Algebra Know-It-ALL

That works! The second original equation comes out like this:

u= 4 t− 3

−17/11= 4 × 4/11 − 3

−17/11= 16/11 − 3

−17/11= 16/11 − 33/11

−17/11= (16 − 33)/11

−17/11=−17/11

That checks out as well. We can be confident that we’ve found the correct solution to the original two-
by-two linear system.

Here’s a challenge!

Derive a general formula using double elimination that solves the following two-by-two linear system for
the variables x and y in terms of the constants a through f. Here are the equations:

ax+by=c

and

dx+ey=f

Solution

First, let’s cause x to disappear so we can solve for y. To do this, we must get coefficients for x that have the
same absolute value but opposite sign in the two equations. Let’s multiply the first equation through by d, and
multiply the second equation through by −a. When we add the resulting equations in their entirety, we get

dax+dby=dc

−adx−aey=−af

dby−aey=dc−af

The terms dax and −adx are additive inverses, so when we add them, they vanish. Now, we can invoke the
distributive law “backward” in the left side of this result to get

(db−ae)y=dc−af

We can solve for y if we divide through by (db−ae), assuming that (db−ae)≠ 0:

y= (dc−af )/(db−ae)

Now let’s cause y to disappear so we can solve for x. We multiply the first original equation through by e, and
multiply the second equation through by −b. When we add the resulting equations in their entirety, we get

eax+eby=ec

−bdx−bey=−bf

eax−bdx=ec−bf

Double Elimination 257