System of Equations
System-of-equations questions in Calculation League use the following formats:
Easy (2 variables):
Ax + By = C
Dx + Fy = G
Advanced (3 variables):
Ax + By + Cz = D
Fx + Gy + Hz = J
Kx + Ly + Mz = N
Expert (usually 3 variables, sometimes 4 variables):
Ax + By + Cz + Dw = F
Gx + Hy + Jz + Kw = L
Mx + Ny + Pz + Qw = R
Sx + Ty + Uz + Vw = ß
These question formats are intentionally artificial: the solutions are generated in advance so that each variable is an integer between 1 and 30.
Example 1 (Easy)
214x + 5y = 677
4x + 2y = 26
On easy level, some problems can be solved immediately using the 1-to-30 constraint.
In the first equation, if x = 1 or x = 2, then y > 30. If x ≥ 4, then y becomes negative. Neither is allowed. Therefore, x must be 3.
Now substitute x = 3:
(214)(3) + 5y = 677
642 + 5y = 677
5y = 35
y = 7
Answer: 3 7
Example 2 (Easy)
765x + 658y = 29316
38x + 85y = 2084
This one is dramatically harder. Here are three practical approaches.
Option A: Standard elimination, but with estimation
If we eliminate x by multiplying:
Multiply the second equation by 765
Multiply the first equation by 38
You get two “y-only” equations:
(38·658)y = 29316·38
(765·85)y = 2084·765
If you can do the full products, great—but you can also estimate.
Left side coefficients:
38·658 ≈ 25,000
765·85 ≈ 65,000
Difference ≈ 40,000
Right side totals:
29316·38 ≈ 1.11 million
2084·765 ≈ 1.59 million
Difference ≈ 0.48 million
So y ≈ 480,000 / 40,000 ≈ 12.
Now plug y = 12 into the easier equation:
38x + 85(12) = 2084
38x + 1020 = 2084
38x = 1064
x = 1064/38 = 28
Answer: 28 12
Option B: Use the 1–30 bounds to narrow the search
From the second equation:
If x = 30: 38(30) + 85y = 2084 → y ≥ 12
From the first equation:
If y = 30: 765x + 658(30) = 29316 → x ≥ 13
So you can start at y = 12 and plug into the second equation. If you don’t get lucky immediately, subtract 85 repeatedly (or 170 at a time, since 38x is always even) until the remainder is divisible by 38.
A useful “quick sense-check” tactic is comparing weighted averages:
In equation 1, coefficient sum = 765 + 658 = 1423, and RHS = 29316 → average value ≈ 29316/1423 ≈ 20.6
In equation 2, coefficient sum = 38 + 85 = 123, and RHS = 2084 → average value ≈ 2084/123 ≈ 16.9
Since y has the heavier weight in equation 2, and the average there is smaller, it suggests y < x.
Option C: Normalize each equation by its coefficient sum
This is a “more automatic” estimation approach:
Equation 1:
765x + 658y = 29316
Divide by 1423:
0.54x + 0.46y ≈ 20.6
Equation 2:
38x + 85y = 2084
Divide by 123:
0.31x + 0.69y ≈ 16.9
This tends to produce fast, usable approximations when elimination feels too heavy.
Example 3 (Advanced)
58x + 3y + 4z = 387
2x + 59y + 800z = 21811
651x + 8y + 7z = 2922
On advanced level, there are three equations and three variables. This example is reasonable because some coefficients are so dominant that they force tight ranges.
Focus on the middle equation:
2x + 59y + 800z = 21811
Since x and y are 1–30:
2x + 59y ranges from 61 to 1830
So:
21811 − 1830 ≤ 800z ≤ 21811 − 61
19981 ≤ 800z ≤ 21750
So z is between 25 and 27.
Test each:
z = 25 → 2x + 59y = 1811 (not possible under 1–30)
z = 26 → 2x + 59y = 1011 (possible: x = 4, y = 17 works)
z = 27 → 2x + 59y = 211 (possible: x = 17, y = 3 works)
Now check against the other equations: x is clearly not 17.
Answer: 4 17 26
Example 4 (Advanced)
6x + 4y + 264z = 3882
910x + 86y + 9z = 23650
280x + 2y + 3z = 7060
The third equation almost directly gives x:
280x ≤ 7060 ≤ 280x + 2(30) + 3(30)
So x is the integer part of 7060/280 = 25.
Plug x = 25 into equation 2:
910(25) + 86y + 9z = 23650
22750 + 86y + 9z = 23650
86y + 9z = 900
Now 86y is between 86 and 2580, but also:
900 − 9(30) ≤ 86y ≤ 900 − 9(1)
630 ≤ 86y ≤ 891
So y is between 8 and 10.
Try y = 10 → 9z = 40 (no)
Try y = 9 → 86(9) = 774 → 9z = 126 → z = 14 (yes)
Answer: 25 9 14
Example 5 (Advanced)
29x + 93y + 2z = 1219
196x + 84y + 9z = 4854
x + 7y + 18z = 531
This is a harder three-equation system because each equation “features” a different variable.
Step 1: crude ranges using the 1–30 bounds
From equation 1:
1219 − 29(30) − 2(30) ≤ 93y ≤ 1219 − 29(1) − 2(1)
289 ≤ 93y ≤ 1188 → y between 4 and 12 (loose)
From equation 2:
4854 − 84(30) − 9(30) ≤ 196x ≤ 4854 − 84(1) − 9(1)
2064 ≤ 196x ≤ 4761 → x between 11 and 24
From equation 3:
531 − 7(30) − (30) ≤ 18z ≤ 531 − 7(1) − (1)
291 ≤ 18z ≤ 523 → z between 17 and 29
Step 2: tighten ranges by substituting the new ranges back in
You can tighten each variable further by using the updated bounds inside the other equations. After tightening, it becomes practical to focus on the “easiest” equation—often the one where the large coefficient dominates.
Here, equation 1 is best to brute-force lightly by trying y values that the tightened range suggests.
Try y = 6, 7, 8:
29x + 2z = 1219 − 93y
y = 6 → 29x + 2z = 661
y = 7 → 29x + 2z = 568
y = 8 → 29x + 2z = 475
Each has (usually) only one feasible (x, z) pair in 1–30. Testing quickly:
y = 6 gives (x, z) = (21, 26)
Now verify in equation 3:
21 + 7(6) + 18(26) = 21 + 42 + 468 = 531 ✅
Answer: 21 6 26
Option C: normalized “weighted average” method
Divide each equation by the sum of its coefficients to convert it into a weighted-average view:
29x + 93y + 2z = 1219 → sum 124
0.23x + 0.75y + 0.02z ≈ 9.8
196x + 84y + 9z = 4854 → sum 289
0.68x + 0.29y + 0.03z ≈ 16.8
x + 7y + 18z = 531 → sum 26
0.04x + 0.27y + 0.69z ≈ 20.4
This is not “systematic algebra,” but it can quickly point you toward y being small-ish and z being large-ish—then you confirm using one equation precisely.
Example 6 (Expert: 4 variables)
310x + 97y + 537z + 69w = 24621
564x + 48y + 3z + 898w = 37943
2x + 407y + 606z + 67w = 27704
87x + 2y + 528z + 6w = 16129
Solving a 4×4 system mentally in 30 seconds is extremely difficult. In actual match conditions, the most realistic plan is:
Start with the equation with the strongest “dominant coefficient” structure (here the 4th equation).
Use rough mid-range guesses (like 15–16) for non-dominant variables to estimate the dominant one.
Move to another equation where a different variable becomes dominant, and repeat.
Once you have a near-solution, pick one equation and compute it exactly to generate an error term you can correct.
One practical correction approach
Suppose you arrive at an initial estimate:
x = 20, y = 24, z = 28, w = 28
Now compute the 4th equation exactly:
87(20) + 2(24) + 528(28) + 6(28)
= 1740 + 48 + 14784 + 168
= 16740
But the RHS is 16129, so you are 611 too high.
Now treat the needed adjustment as:
87Δx + 2Δy + 528Δz + 6Δw = −611
A very fast observation is:
87 + 528 = 615, and 2 − 6 = −4
So the adjustment:
Δx = −1, Δy = −1, Δz = −1, Δw = +1
changes the total by:
−87 − 2 − 528 + 6 = −611
So the corrected solution is:
x = 19, y = 23, z = 27, w = 29
Answer: 19 23 27 29