3×3 systems: use elimination to reduce to 2×2, not brute substitution

In short

Faced with three equations in three unknowns, do not isolate one variable from one equation and substitute the messy expression into the other two. That path piles fractions on fractions. The clean move is elimination, twice: pick a variable (the one with the friendliest coefficients), kill it once from one pair of equations to get a new equation A in two variables, then kill the same variable from a different pair to get B in the same two variables. Now you are looking at a 2×2 system \{A, B\} — a problem you already know how to solve. Find the two unknowns, then plug back into any of the original three equations to recover the third. This is the manual version of Gaussian elimination, the matrix method you will meet again in CBSE Class 11.

You are sitting with three equations in x, y, z. Maybe a problem about three shopkeepers, or three lines in 3D space, or the currents in three branches of a circuit. You stare at it and your first instinct — the one trained by every two-variable problem you have ever solved — whispers: isolate one variable and substitute.

Resist that instinct. With three equations the substitution route is a trap.

Here is why. If you take x = (\text{stuff in } y \text{ and } z) from equation 1 and plug it into equation 2, you get a new equation in y and z — fine so far. But you must also plug the same messy expression into equation 3. Now you are copying the same fractions twice, and any small mistake (a sign, a denominator) infects both of your remaining equations. The arithmetic snowballs.

Elimination keeps the arithmetic flat.

The recipe

Why this works: every linear combination of two equations is itself a valid equation that the solution must satisfy. So you are allowed to add, subtract, scale, and combine equations as long as you do it consistently. The 2×2 system you produce has the same solution (y, z) as the original 3×3.

Pick the variable to eliminate first. Look at all three equations and choose the variable whose coefficients are easiest to match — usually the one that already has a coefficient of 1 in some equation, or whose coefficients share a small common multiple.
Combine equation 1 and equation 2 to kill that variable. Call the result equation A. It contains only the other two variables.
Combine equation 1 and equation 3 (or equation 2 and equation 3) to kill the same variable again. Call the result equation B. It also contains only those same two variables.
Solve the 2×2 system \{A, B\} by ordinary elimination or substitution. You now have two of the three unknowns.
Back-substitute into any of the original three equations to recover the variable you eliminated.

That is the whole method. The picture below makes the shape of the move obvious.

Two pairings of the three equations, each killing the same variable, leaves you with two equations in two unknowns — a problem you have already mastered.

Worked example 1 — clean elimination

Solve

x + y + z = 6, \qquad x + 2y + 3z = 14, \qquad x + 4y + 9z = 36.

Step 1 — choose what to eliminate. Every equation has x with coefficient 1. That is the friendliest possible setup, so kill x first.

Step 2 — eliminate x from the first pair. Subtract eq 1 from eq 2:

(x + 2y + 3z) - (x + y + z) = 14 - 6 \implies y + 2z = 8. \quad (A)

Step 3 — eliminate x from a second pair. Subtract eq 1 from eq 3:

(x + 4y + 9z) - (x + y + z) = 36 - 6 \implies 3y + 8z = 30. \quad (B)

Why subtract from the same equation both times: it keeps the bookkeeping symmetric. You could also subtract eq 2 from eq 3 — any two independent combinations work.

Step 4 — solve the 2×2 system. Multiply (A) by 3: 3y + 6z = 24. Subtract from (B):

(3y + 8z) - (3y + 6z) = 30 - 24 \implies 2z = 6 \implies z = 3.

Plug into (A): y + 6 = 8 \implies y = 2.

Step 5 — back-substitute. Use eq 1: x + 2 + 3 = 6 \implies x = 1.

Solution: (x, y, z) = (1, 2, 3). Check eq 3: 1 + 8 + 27 = 36. ✓

Notice how flat the arithmetic stayed. No fractions appeared until you might have chosen to introduce them, and you did not have to.

Worked example 2 — mixed signs

Solve

x + y + z = 9, \qquad 2x + y - z = 3, \qquad x - y + 2z = 7.

Pick the eliminator. The coefficients of z are 1, -1, 2. The first two are negatives of each other, so adding eq 1 and eq 2 will kill z in one stroke.

Eliminate z from eq 1 + eq 2:

3x + 2y = 12. \quad (A)

Eliminate z from a second pair. Multiply eq 1 by 2 and subtract eq 3:

2(x + y + z) - (x - y + 2z) = 18 - 7 \implies x + 3y = 11. \quad (B)

Solve the 2×2. Multiply (A) by 1 and (B) by 3 to align y:

3x + 2y = 12, \qquad 3x + 9y = 33.

Subtract: 7y = 21 \implies y = 3. Plug back into (A): 3x + 6 = 12 \implies x = 2.

Back-substitute into eq 1: 2 + 3 + z = 9 \implies z = 4.

Solution: (2, 3, 4). Check eq 3: 2 - 3 + 8 = 7. ✓

Two passes of elimination, one easy 2×2, done. The whole solution fits on half a page of your notebook.

Worked example 3 — what brute substitution would cost you

Take the system from example 1 and try the substitution route instead, just to feel the difference.

From eq 1, isolate x: x = 6 - y - z.

Substitute into eq 2:

(6 - y - z) + 2y + 3z = 14 \implies y + 2z = 8.

Fine — same as (A) above.

Substitute into eq 3:

(6 - y - z) + 4y + 9z = 36 \implies 3y + 8z = 30.

Also fine — same as (B).

So in this particular example the substitution route landed on the same two reduced equations. That is because the coefficient of x was 1 in eq 1, which is the lucky case where substitution and elimination cost almost the same.

Now try it on example 2, where eq 2 starts with 2x. From eq 1, x = 9 - y - z. Plug into eq 2:

2(9 - y - z) + y - z = 3 \implies 18 - 2y - 2z + y - z = 3 \implies -y - 3z = -15 \implies y + 3z = 15.

Plug the same x = 9 - y - z into eq 3:

(9 - y - z) - y + 2z = 7 \implies -2y + z = -2.

Compare with the elimination route (which gave 3x + 2y = 12 and x + 3y = 11 — small numbers, no copying of bracketed expressions). The substitution route forced you to expand 2(9 - y - z) and (9 - y - z) once each, producing four sign flips per equation. With three-digit constants or fractional coefficients those expansions become the place where exam marks die.

Why elimination is safer: each step is a single linear combination of two existing equations. You never copy a multi-term expression into another equation. Substitution forces that copy twice.

When substitution is fine

There is one situation where reaching for substitution is reasonable: a variable already appears alone in one equation, like z = 2 or x = y + 1. Then substituting it everywhere is just bookkeeping — no expansion, no fractions. In every other case, eliminate.

Connection to matrices and Gaussian elimination

When you reach CBSE Class 11 and meet matrices, you will see the same procedure dressed up in new clothes. Writing the system as

\begin{pmatrix} a_1 & b_1 & c_1 & | & d_1 \\ a_2 & b_2 & c_2 & | & d_2 \\ a_3 & b_3 & c_3 & | & d_3 \end{pmatrix}

and applying row operations R_2 \to R_2 - \tfrac{a_2}{a_1} R_1 and R_3 \to R_3 - \tfrac{a_3}{a_1} R_1 is exactly the move you just performed by hand: eliminate the first variable from rows 2 and 3 using row 1. The matrix layout is just a tidier scoreboard for the same arithmetic. Once you internalise the pen-and-paper version, Gaussian elimination is no surprise — it is the formal name for the habit you already have.

A quick checklist for the exam hall

Three equations? Look for a variable with friendly coefficients (a 1, a -1, or a small common factor). That is your first eliminator.
Eliminate it from two different pairs of equations. Write down equations A and B, both in the same two remaining variables.
Solve the 2×2 the way you always have.
Back-substitute into the simplest of the original three equations.
Always verify by plugging (x, y, z) into the equation you did not use for back-substitution.

Three equations, three unknowns — but only two new equations to write down. That is the saving the elimination route gives you, and the reason it is the default for every solver from a Class 10 student to a numerical-linear-algebra library.