Mathematical Equations
What this question type is
A small system of linear equations using letters as unknowns. Every letter equals a unique whole number between 1 and 20, and there's exactly one solution.
How to solve it
Find the equation you can act on first — usually one that defines a letter in terms of another (like "B = 2 × A"). Substitute it into a second equation to reduce the number of unknowns, solve for one letter, then back-substitute to get the rest.
Worked example
Given B = 2 × A and B + A = 9: substitute the first into the second → 2A + A = 9 → 3A = 9 → A = 3, so B = 6.
Try it yourself
0 of 20 solved- B = 2 × A
- B + A = 9
Why
Substitute B with 2 × A in the second equation: 2A + A = 9, so 3A = 9 and A = 3. Then B = 2 × 3 = 6.
- A + 4 = B
- A + B = 14
Why
From the first equation B = A + 4. Substitute into the second: A + (A + 4) = 14, so 2A = 10 and A = 5. Then B = 9.
- C + D − A = 5
- 2 × C = D
- 7 − C = A
- 3 × C − 1 = B
Why
Solve from C. Try C = 3: D = 2 × 3 = 6, A = 7 − 3 = 4, B = 3 × 3 − 1 = 8. Check the first equation: 3 + 6 − 4 = 5 ✓. So A=4, B=8, C=3, D=6.
- B = 3 × A
- B + A = 12
Why
Substitute B with 3 × A: 3A + A = 12, so 4A = 12 and A = 3. Then B = 3 × 3 = 9.
- B + D − C = 11
- 2 × A + 1 = B
- 3 × A = D
- A + 6 = C
Why
Solve from A. The last three equations give B = 2A + 1, D = 3A, and C = A + 6. Substitute into the first equation: (2A + 1) + 3A − (A + 6) = 11, so 4A − 5 = 11, giving 4A = 16 and A = 4. Then B = 2 × 4 + 1 = 9, D = 3 × 4 = 12, and C = 4 + 6 = 10.
- C − D = 13
- A = D + 2
- B = 2 × A − 1
- C = B + 3
Why
Follow the chain from D. From the second equation, A = D + 2. Substitute into the third: B = 2 × (D + 2) − 1 = 2D + 3. Substitute into the fourth: C = (2D + 3) + 3 = 2D + 6. Substitute into the first equation: (2D + 6) − D = 13, so D + 6 = 13 and D = 7. Then A = 7 + 2 = 9, B = 2 × 9 − 1 = 17, and C = 17 + 3 = 20.
- B = 2 × A
- B + A = 9
Why
Substitute B with 2 × A in the second equation: 2A + A = 9, so 3A = 9 and A = 3. Then B = 2 × 3 = 6.
- A + 6 = B
- A + B = 16
Why
From the first equation, B = A + 6. Substitute into the second: A + (A + 6) = 16, so 2A = 10 and A = 5. Then B = 11.
- B − 6 = A
- B + A = 16
Why
From the first equation, A = B − 6. Substitute into the second: B + (B − 6) = 16, so 2B = 22 and B = 11. Then A = B − 6 = 5.
- 3 × C = A
- A + C = 8
- 2 × A + 2 × C = B
Why
The first equation gives A = 3C. Substitute into the second: 3C + C = 8, so 4C = 8 and C = 2. Then A = 3 × 2 = 6, and B = 2 × 6 + 2 × 2 = 16.
- 3 × C = A
- A + C = 8
- 3 × A − C = B
Why
The first equation gives A = 3C. Substitute into the second: 3C + C = 8, so 4C = 8 and C = 2. Then A = 3 × 2 = 6, and B = 3 × 6 − 2 = 16.
- 2 × B = C
- C − B = 2
- A = C + B
Why
The first equation gives C = 2B. Substitute into the second: 2B − B = 2, so 1B = 2 and B = 2. Then C = 2 × 2 = 4, and A = C + B = 6.
- B = 2 × A
- A + B = 6
- C = A + B
Why
Substitute B with 2 × A in the second equation: 2A + A = 6, so 3A = 6 and A = 2. Then B = 2 × 2 = 4, and C = A + B = 6.
- C − D = 7
- A = D + 1
- B = 2 × A − 1
- C = B + 2
Why
Follow the chain from D. From the second equation, A = D + 1. Substitute into the third: B = 2 × (D + 1) − 1 = 2D + 1. Substitute into the fourth: C = (2D + 1) + 2 = 2D + 3. Substitute into the first equation: (2D + 3) − D = 7, so D + 3 = 7 and D = 4. Then A = 4 + 1 = 5, B = 2 × 5 − 1 = 9, and C = 9 + 2 = 11.
- A − B + C − D = −6
- B = A + 2
- D = 4 × A
- C = D − B
Why
Express B, D and C in terms of A: B = A + 2, D = 4 × A, and C = D − B = 4A − (A + 2) = 3A − 2. Substitute all three into the first equation: A − (A + 2) + (3A − 2) − 4A = −6. The A terms combine to −A, so this simplifies to −A − 4 = −6, giving A = 2. Then B = 2 + 2 = 4, D = 4 × 2 = 8, and C = 8 − 4 = 4.
- C + D − A = 0
- 4 × C = D
- 12 − C = A
- 3 × C − 3 = B
Why
The information given in equations two and three for A and D can be inserted into the first equation, so it can be solved for C: C + 4C − (12 − C) = 0. This gives 6C − 12 = 0, so 6C = 12 and C = 2. This can be inserted into the other equations: 4 × 2 = D or D = 8, 12 − 2 = A or A = 10, and 3 × 2 − 3 = B or B = 3.
- A + B − C + D = 9
- B = A − 2
- C = 2 × B + 1
- D = C − 1
Why
Express B, C and D in terms of A: B = A − 2, C = 2 × (A − 2) + 1 = 2A − 3, and D = C − 1 = 2A − 4. Substitute all three into the first equation: A + (A − 2) − (2A − 3) + (2A − 4) = 9. The A terms combine to 2A, so this simplifies to 2A − 3 = 9, giving A = 6. Then B = 6 − 2 = 4, C = 2 × 4 + 1 = 9, and D = 9 − 1 = 8.
- B + D = 10
- C = 4 × D
- A = C + 1
- B = A − 1
Why
Follow the chain from D: C = 4D. Substitute into the third equation: A = 4D + 1. Substitute into the fourth: B = (4D + 1) − 1 = 4D. Substitute into the first equation: (4D) + D = 10, so 5D = 10 and D = 2. Then C = 4 × 2 = 8, A = 8 + 1 = 9, and B = 9 − 1 = 8.
- D − B = 5
- A = 3 × B
- C = A + 3
- D = C − 2
Why
Follow the chain from B: A = 3B. Substitute into the third equation: C = 3B + 3. Substitute into the fourth: D = (3B + 3) − 2 = 3B + 1. Substitute into the first equation: (3B + 1) − B = 5, so 2B + 1 = 5 and B = 2. Then A = 3 × 2 = 6, C = 6 + 3 = 9, and D = 9 − 2 = 7.
- A + D = 12
- A = 3 × C
- B = A + 3
- D = B − 3
Why
Follow the chain from C: A = 3C. Substitute into the third equation: B = 3C + 3. Substitute into the fourth: D = (3C + 3) − 3 = 3C. Substitute into the first equation: 3C + (3C) = 12, so 6C = 12 and C = 2. Then A = 3 × 2 = 6, B = 6 + 3 = 9, and D = 9 − 3 = 6.