Three lessons into Three-Variable Systems and a First Look Beyond, and we have yet to write down a single ordered triple that came out of an actual solve. We have recognized triples, tested them, eliminated a variable twice, and scaled coefficients until they cancelled — and every one of those lessons ended in the same unsatisfying spot: a neat two-variable system on the page, with the real answer still hiding behind it.
This fourth lesson finishes the job. Someone hands us the reduced system; we solve it, travel back to an original equation to recover the variable we threw away, and confirm the resulting triple in all three original equations. There is no reduction to perform today. This lesson is entirely about the last mile.
The Last Mile of a Three-Variable Solve
Think of solving a three-variable system as a round trip. On the way out, elimination takes you down a level: three unknowns collapse into two. On the way back, substitution lifts you up again, because two known values are enough to recover the one you set aside. The reduced system is the turning point — the farthest point from home, and the place where the return journey begins.
That picture fixes the order of the finishing steps:
Solve the reduced two-variable system for its two variables.
Back-substitute both values into one original equation to find the eliminated variable.
Verify the complete triple in all three original equations.
Steps 1 and 2 use tools you already own from the substitution and elimination courses. Step 3 is where three-variable work genuinely differs, and it is the step we will spend the most care on.
Solving the Reduced System
Let's pick up precisely where the previous lesson left off. From the system
This is an ordinary two-variable system, so any method works. Both equations carry a plain y with the same sign, which makes subtraction the cheapest move:
−​y+10z=32y−2z=−412z=36​​
So z=3. Substituting that into y−2z=−4 gives y−6=−4, hence y=2. Two of the three values are now locked in — and notice that the reduced system has told us everything it possibly can.
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Back-Substituting for the Third Variable
Choosing a Convenient Equation for Back-Substitution
We eliminated x, so x is the value still missing, and no amount of staring at the reduced system will produce it: x simply is not in there anymore. To recover it we must return to an original equation, one that still contains an x term. Using equation (1):
2x+3(2)+3(3)=17
Simplify the known part first: 2x+6+9=17, so 2x=2 and x=1. Our candidate solution is the ordered triple (1,2,3).
One caution worth repeating from Lesson 1: the triple is written in the order (x,y,z), never in the order the values happened to appear. We found z first, then y, then x — and we still report (1,2,3).
Any of the three original equations would have produced x=1, but they do not all cost the same effort. Before substituting, glance down the column of the missing variable and at the size of the surrounding numbers:
Equation
Coefficient of x
Work required
(1)Â 2x+3y+3z=17
2
Small products, one easy division
(2) 4x+5y−4z=2
4
Larger products, a negative term to track
(3)Â 6x+11y+5z=43
6
Largest numbers of the three
Equation (1) wins on every count, which is why we used it. The best equation to back-substitute into is the one where the missing variable has the coefficient with the smallest absolute value — ideally 1 or −1 — and where the remaining numbers stay manageable. If a coefficient of 1 or −1 appears anywhere in that column, that equation is almost always the right choice, since it removes the division step entirely.
Now comes the step learners most often skip. We used equation (1) to findx, so equation (1) is guaranteed to work; checking it proves nothing at all. A genuine verification pushes (1,2,3) through all three original equations:
Think of a nutritionist mixing three foods to hit a calorie target, a protein target, and a budget. Getting the calories right means nothing if the protein is off — the mix only works when every requirement holds at once, and each original equation is one of those requirements.
Checking all three is also a free error detector. If equation (2) or (3) fails, the mistake lives upstream in the reduction or in the solving, and you know to go back rather than trust a wrong triple. Note that we check the original equations, not the reduced ones: a scaling slip during reduction would be invisible to the very equations it corrupted.
Signs are where good work quietly falls apart, so let's walk the same three steps with a triple that is not all positive. Suppose the original system is
The first equation rearranges to z=3y, which is about as friendly a substitution as we could hope for. Putting that into the second gives y−2(3y)=5, so −5y=5 and y=−1. Then z=3(−1)=−3.
Equation (1) has a coefficient of 1 on x, so it is the obvious landing spot: x+(−1)+(−3)=−2 becomes x−4=−2, so x=2. Verifying the triple (2,−1,−3):
Watch what happened in equation (2): the term −y became −(−1)=+1. Wrapping every substituted value in parentheses before simplifying is the single habit that prevents most of these slips.
The arithmetic on the last mile is lighter than in the reduction, which is exactly why attention tends to drift. These are the errors worth guarding against:
Back-substituting into the reduced system. Those equations no longer contain the eliminated variable, so they can never reveal it. Return to an original equation.
Verifying in only one equation. Checking the equation you used for back-substitution is circular reasoning. All three must be tested.
Reporting values in the order you found them. Solve order and triple order are unrelated; always write (x,y,z).
Dropping a negative during substitution. Parentheses around every value, every time.
Stopping at two values. A three-variable system is not solved until all three numbers exist and have been checked together.
In this lesson we completed the round trip. Starting from a reduced two-variable system, we solved for its two variables, picked the friendliest original equation and back-substituted to recover the eliminated one, assembled the results into an ordered triple in (x,y,z) order, and confirmed that triple in every original equation. Small numbers, big responsibility: this is the stage where a solution either becomes trustworthy or quietly goes wrong.
The practice ahead walks that last mile alongside you. You will fill in a guided finish, recover a missing third value on your own, complete two full solutions — one of them packed with negatives — and decide for yourself which original equation makes back-substitution easiest. Let's bring these triples home.
