Finding the Unknown · Lesson 3 of 13
Trial, Error, and Preserving Equality
“Move from sensible trials to operations that keep an equation balanced.”
• Use a trial table to search for a solution. • Explain why the same operation on both sides preserves equality. • Use inverse operations to undo addition and multiplication. • Recognise when trial and error is inefficient.
Try a value and compare
Trial and error means choosing a value, substituting it, and seeing whether the result is too small, too large, or exactly right. It becomes useful reasoning when each trial guides the next one. For 2n + 1 = 99, increasing n makes the left side larger, so a value giving too small a result tells us to try a larger n.
| Trial n | 2n + 1 | Comparison with 99 |
|---|---|---|
| 5 | 11 | Too small |
| 10 | 21 | Too small |
| 30 | 61 | Too small |
| 40 | 81 | Too small |
| 50 | 101 | Too large |
| 49 | 99 | Equal |
Problem
Find the solution of 2n + 1 = 99 using trials.
- 1.n = 40 gives 81, while n = 50 gives 101, so the required value lies between them.
- 2.n = 49 gives 99. A change of 1 in n changes 2n + 1 by 2.
- 3.Every larger n gives a larger result, and every smaller n gives a smaller result. Thus no second value gives 99.
Trial and error can be awkward when the solution is fractional. In 5x − 4 = 7, x = 2 gives 6 and x = 3 gives 11. Trying only whole numbers would miss x = 2.2. We therefore need a method that follows equality itself and does not depend on guessing a convenient kind of answer.
Make equal changes to equal quantities
If two amounts are equal, adding the same number to both leaves them equal. Subtracting the same number also works. Multiplying both sides by the same number preserves equality, and dividing both sides by the same non-zero number preserves it too. These rules let us simplify an equation without favouring one side.
| Starting equality | Operation on both sides | New equality |
|---|---|---|
| a = b | Add c | a + c = b + c |
| a = b | Subtract c | a − c = b − c |
| a = b | Multiply by c | ac = bc |
| a = b | Divide by c, with c ≠ 0 | a/c = b/c |
When solving, choose operations that can be undone. Addition and subtraction undo each other; multiplication and division by a non-zero number undo each other. Multiplying by zero makes any two sides zero and loses information about the original unknown. Division by zero is undefined. Neither is a useful solving step.
Operations that undo each other, such as adding 8 and subtracting 8, or multiplying by 7 and dividing by 7.
Problem
E + 88 = 13353. Find E, then compare with E = 14593 − 1459 + 145 − 14.
- 1.Subtract 88 from both sides: E + 88 − 88 = 13353 − 88, so E = 13265.
- 2.The longer expression gives 14593 − 1459 = 13134; +145 = 13279; −14 = 13265.
- 3.Check the equation: 13265 + 88 = 13353. Inverse operations reveal the missing number directly.
Problem
P × 7 = 580888. Find P; compare with P = 23 × 41 × 11 × 8.
- 1.Divide both sides by 7: P = 580888 ÷ 7 = 82984.
- 2.For the product, 23 × 41 = 943; 943 × 11 = 10373; 10373 × 8 = 82984.
- 3.Check: 82984 × 7 = 580888. The multiplier 7 can be undone because it is non-zero.
Keep signs and fractions meaningful
An operation is determined by the number being added or multiplied, including its sign. Subtracting a negative number adds its positive opposite. To undo multiplication by a non-zero fraction, multiply by its reciprocal: the fraction obtained by exchanging numerator and denominator. Their product is 1.
Problem
E − (−67) = 7091. Find E, also given by 12345 − 5432 + 135 − 24.
- 1.Subtracting −67 means adding 67, so E + 67 = 7091.
- 2.Subtract 67 from both sides: E = 7024.
- 3.The longer expression gives 6913 + 135 − 24 = 7048 − 24 = 7024. Check: 7024 − (−67) = 7091.
Problem
F × 8/9 = 94080/1017. Find F.
- 1.Multiply both sides by 9/8. The left becomes F because (8/9) × (9/8) = 1.
- 2.F = (94080/1017) × (9/8). Since 1017 = 9 × 113, simplify to F = 11760/113.
- 3.This agrees with (35/113) × 24 × 14, whose numerator is 35 × 336 = 11760. Multiplying the answer by 8/9 returns 94080/1017.
Problem
Solve 5x − 4 = 7.
- 1.Add 4 to both sides: 5x − 4 + 4 = 7 + 4, so 5x = 11.
- 2.Divide both sides by 5: x = 11/5 = 2.2.
- 3.Check: 5 × 2.2 − 4 = 11 − 4 = 7. A fractional solution is perfectly valid here.
Do not alter just the left side of an equation. A useful simplification must apply the same operation to the entire right side too. Never divide by zero or use multiplication by zero to recover an unknown.
Quiz
A trial gives a value too small for 2n + 1 = 99. What should you try next?
Which operation undoes adding 88?
What operation undoes multiplication by 8/9?
Which step preserves x + 6 = 14 and helps isolate x?
What is the solution of 5x − 4 = 7?
Practice Problems
- Make a trial table for 3x + 2 = 23, using at least three values before identifying the solution.
- For x − 19 = 45, state the inverse operation, apply it to both sides, and check.
- For 6p = 102, explain why division by 6 isolates p.
- Solve a − (−12) = 31 and verify the sign in your check.
- Solve b × 3/5 = 12 using a reciprocal. Explain why the left side becomes b.
- Explain why trying only whole numbers cannot establish that 4t = 7 has no solution.
- Write a short explanation of why multiplying both sides by zero hides the original information.
Key Takeaways
• Each trial must be checked by substitution. • Use trials to guide a search rather than guessing blindly. • Apply the same operation to both entire sides. • Inverse operations undo the changes around the unknown. • Divide only by non-zero numbers, and allow fractional solutions.