Reduce it to a problem you have already solved: two equations, two unknowns.
Solve the system
x + y + z = 6
2x − y + 3z = 9
x + 2y − z = 2
Kill y first: adding equations 1 and 2 gives 3x + 4z = 15, and equation 3 minus twice equation 1 gives x + 3z = 10. Those two solve to z = 3 and x = 1, leaving y = 2. Check in equation 2: 2 − 2 + 9 = 9 ✓ Every choice adds up to 6, so equation 1 alone proves nothing — A (2, 1, 3) passes equation 1 but fails equation 2: 4 − 1 + 9 = 12, not 9.
| Stage | What you get |
|---|---|
| pair one | two equations in x and z |
| pair two | solve that 2 × 2 system |
| climb back | substitute for the third letter |
Pick the easiest letter to remove. Here y already has coefficients +1, −1 and +2 — adding E1 and E2 deletes it with no multiplying at all.
x + y + z = 6
+ 2x − y + 3z = 9
3x 0y + 4z = 15
Call this E4.
E3 has 2y, so double E2 first to make the y terms cancel.
4x − 2y + 6z = 18
+ x + 2y − z = 2
5x 0y + 5z = 20
Divide by 5: x + z = 4. Call this E5.
Removing y once and z once leaves two equations that share no unknowns, and you are no further forward. Both eliminations must target y.
x = 4 − z
E5 is easy to rearrange
3(4 − z) + 4z = 15
substitute into E4
12 − 3z + 4z = 15
expand
z = 3
one letter down
x = 4 − 3 = 1
back into E5
Use the simplest original equation — E1.
x + y + z = 6
E1
1 + y + 3 = 6
put x and z in
y = 2
all three found
A three-variable question wants an ordered triple, and sometimes only their sum. Finding z and stopping is the standard trap — the working is done, so spend the ten seconds.
| What you hit | Means |
|---|---|
| one triple | the three planes meet at a point |
| 0 = 5, a contradiction | no solution — the system is inconsistent |
| 0 = 0, always true | infinitely many solutions — one equation added nothing |
Each equation is a plane in three dimensions. A unique solution means all three planes pass through one point; a false statement means they never do.
Sometimes the question wants only x + y + z, and adding all three equations gives it directly. Read what is asked before starting the eliminations — occasionally the whole system is unnecessary.
Solve 2x + y = 5, y − z = 4 and x + 3z = −7.
y = 5 − 2x
E1 has no z: use it to replace y
(5 − 2x) − z = 4 → 2x + z = 1
into E2: now x and z only
x = −7 − 3z → 2(−7 − 3z) + z = 1 → −5z = 15
E3 already has no y
z = −3, x = 2, y = 1 → (2, 1, −3)
check E3: 2 − 9 = −7 ✓
Different: a missing letter means that equation has already done an elimination for us — substitution is quicker than adding pairs.
Solve x + y + z = 4, x − y + 2z = 1 and 2x + 3z = 7.
E1 + E2: 2x + 3z = 5
y removed
but E3 says 2x + 3z = 7
subtract: 0 = 2
no solution
the system is inconsistent
Different: the same left side is asked to equal two numbers. Had E3 ended in = 5, it would add nothing and there would be infinitely many solutions.
When every letter cancels, there is nothing left to solve for. A false line means no triple works at all; do not try to rescue a value from it.
Pair sums: add all three. If a + b = 7, b + c = 10 and a + c = 9, find a, b and c. Every letter appears exactly twice.
2a + 2b + 2c = 26 → a + b + c = 13
add the three equations, halve
c = 13 − (a + b) = 13 − 7 = 6
subtract the pair without c
a = 13 − 10 = 3, b = 13 − 9 = 4
the same move twice more
A parabola through three points. Find y = ax² + bx + c through (0, −1), (1, 2) and (−1, −2). Each point gives one equation in a, b and c.
(0, −1): c = −1
x = 0 kills a and b — use it first
(1, 2): a + b − 1 = 2 → a + b = 3
x² = 1, x = 1
(−1, −2): a − b − 1 = −2 → a − b = −1
x² = 1, x = −1
2a = 2 → a = 1, b = 2
add, then back-substitute
y = x² + 2x − 1
(1): 1 + 2 − 1 = 2 ✓ (−1): 1 − 2 − 1 = −2 ✓
The points supply the x and y. Substitute them as numbers, and remember x² from a negative x is positive: at x = −1 the a-term is +a, not −a.
Three kinds of coins. 13 coins of ₱1, ₱5 and ₱10 are worth ₱61, and there are twice as many ₱1 coins as ₱5 coins. Let the counts be a, b and c.
a + b + c = 13, a + 5b + 10c = 61, a = 2b
count, value (count × face value), the link
3b + c = 13 and 7b + 10c = 61
put a = 2b into both: now two letters
c = 13 − 3b → 7b + 130 − 30b = 61 → b = 3
a = 6, c = 4; 6 + 15 + 40 = 61 ✓
Full three-variable systems are slow, so the exam often wants a shortcut: a sum like x + y + z (add everything), one letter only, or a word problem with three kinds of coins or three numbers. Check whether a substitution like b = 2c turns it into a two-variable system first.
A slip in one elimination can still satisfy two of the equations.
Solve
x + y + z = 12
2x − y + z = 6
x + 2y − 3z = −8
Adding equations 1 and 2 removes y and leaves 3x + 2z = 18; equation 3 minus twice equation 1 gives x + 5z = 32. Solving gives z = 6 and x = 2, so y = 4. Verify in equation 3: 2 + 8 − 18 = −8 ✓ Always eliminate the same variable from both pairings.
