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Lesson 9 of 13

Finding the Unknown · Lesson 9 of 13

Mind the Mistake, Mend the Mistake

“Find the first invalid step and repair the reasoning behind an answer.”

Learning Objectives

• Identify the first step that fails to preserve equality. • Correct sign, coefficient, bracket, and division mistakes. • Distinguish a valid unusual step from an invalid one. • Verify repaired solutions in the original equation.

Audit the step, not the appearance

A solution chain can look tidy and still be wrong. Compare each line with the one immediately before it and ask what operation was applied to both sides. The first line that cannot be justified is the first error. Later lines may follow correctly from that mistaken line, but they are solving a changed equation.

Do not judge a step only by whether it matches your preferred method. For example, 7 − 8z = 5 becoming 8z = 7 − 5 is valid: add 8z to both sides and subtract 5. What matters is equality, not whether a term seems to move in a familiar direction.

An added number gets the wrong sign

Problem
A learner writes 4x + 6 = 10 → 4x = 10 + 6 → 4x = 16 → x = 4. Repair it.

  1. 1.The first change is invalid. Removing +6 from the left requires subtracting 6 from both sides.
  2. 2.Correct chain: 4x = 10 − 6 = 4, then x = 1.
  3. 3.Check: 4 × 1 + 6 = 10. The proposed x = 4 gives 22, so it fails the original equation.
A factor is not an added term

Problem
Inspect 7 − 8z = 5 → 8z = 7 − 5 → 8z = 2 → z = 4.

  1. 1.The first two changes are valid. The first error is changing 8z = 2 to z = 4.
  2. 2.Divide both sides by 8: z = 2/8 = 1/4.
  3. 3.Check: 7 − 8 × (1/4) = 7 − 2 = 5. The coefficient must be undone by division.
Dividing a sum means dividing every term

Problem
Inspect 2v − 4 = 6 → v − 4 = 6 − 2 → v − 4 = 4 → v = 8.

  1. 1.The first change is invalid: the coefficient 2 was neither divided from every term nor removed by an equal operation.
  2. 2.A clean repair is to add 4: 2v = 10, then divide by 2: v = 5.
  3. 3.Alternatively divide every term of the original by 2: v − 2 = 3, then add 2.
  4. 4.Check: 2 × 5 − 4 = 6. Leaving −4 unchanged while dividing 2v changes the equation.

Unknown terms keep their letters

A term such as 4w is four copies of w, not the ordinary number 4. It can be combined with 15w because both contain the same unknown part. When moving an added unknown term through a justified subtraction, its sign changes just as an ordinary added term’s sign does. Keep the letter attached until the unknown is isolated.

Two terms receive wrong signs

Problem
Inspect 5z + 2 = 3z − 4 → 5z + 3z = −4 + 2 → 8z = −2 → z = −2/8.

  1. 1.The first change is invalid. To collect z terms, subtract 3z; to remove +2, subtract 2.
  2. 2.Correct chain: 5z − 3z = −4 − 2, so 2z = −6 and z = −3.
  3. 3.Check: 5(−3) + 2 = −13 and 3(−3) − 4 = −13.
Do not treat 4w as 4

Problem
Inspect 15w − 4w = 26 → 15w = 26 + 4w → 15w = 30 → w = 2.

  1. 1.The first change is valid because it adds 4w to both sides. The next change is invalid: w is still unknown, so 26 + 4w is not 30.
  2. 2.Combine like terms in the original instead: 11w = 26. Divide by 11: w = 26/11.
  3. 3.Check: (15 − 4) × (26/11) = 26.
A constant must be divided too

Problem
Inspect 3x + 1 = −12 → x + 1 = −12/3 → x + 1 = −4 → x = −5.

  1. 1.The first change is invalid. Dividing by 3 would give x + 1/3 = −4.
  2. 2.An easier repair is subtract 1 first: 3x = −13. Divide by 3: x = −13/3.
  3. 3.Check: 3(−13/3) + 1 = −13 + 1 = −12.

Respect the whole bracket

A bracket groups terms before an outside multiplication. Distributing a positive or negative factor requires multiplying every term, including its sign. Dividing a side that contains a sum also affects every term. Writing the expanded or divided expression in full is a useful safeguard before doing arithmetic.

Remove the actual constant term

Problem
Inspect 4(4q + 2) = 50 → 4(4q) = 50 − 2 → 16q = 48 → q = 3.

  1. 1.The first change is invalid: the added 2 inside the bracket contributes 4 × 2 = 8 to the left side.
  2. 2.Expand correctly: 16q + 8 = 50. Subtract 8: 16q = 42. Divide by 16: q = 21/8.
  3. 3.Check: 4(4 × 21/8 + 2) = 4(21/2 + 2) = 4 × 25/2 = 50.
A negative times a negative is positive

Problem
Inspect −2(3 − 4x) = 14 → −6 − 8x = 14.

  1. 1.The expansion is the first error. Multiplying −2 by −4x gives +8x, so the correct equation is −6 + 8x = 14.
  2. 2.Add 6: 8x = 20. Divide by 8: x = 5/2.
  3. 3.Check: −2(3 − 4 × 5/2) = −2(3 − 10) = −2(−7) = 14.
Divide the complete right side

Problem
Inspect 3(7y + 4) = 9 + 5y → 7y + 4 = 3 + 5y → 7y − 5y + 4 = 3 → 2y = 4 − 3 → y = 1/2.

  1. 1.The first change is already invalid: division by 3 should give 7y + 4 = 3 + 5y/3. The later step 2y = 4 − 3 also reverses the subtraction: removing +4 would require 3 − 4.
  2. 2.Expand the original: 21y + 12 = 9 + 5y. Subtract 5y: 16y + 12 = 9.
  3. 3.Subtract 12: 16y = −3. Divide by 16: y = −3/16.
  4. 4.Check: LHS = 3(−21/16 + 64/16) = 129/16; RHS = 144/16 − 15/16 = 129/16.

Some chains are completely correct

Error checking also means recognising when there is no error. Dividing an entire equation by a common factor can produce a less familiar but valid line. State why it works rather than inventing a correction. The following comparisons extend the same checking method to three more chains.

Proposed chainDiagnosisCorrect result or justification
6x + 9 = 66 → x + 9 = 11 → x = 2First change divides only part of the leftSubtract 9: 6x = 57; x = 19/2; check 57 + 9 = 66
14y + 24 = 36 → 7y + 12 = 18 → 7y = 6 → y = 6/7Every change is validDivide all terms by 2, subtract 12, divide by 7; check 12 + 24 = 36
4x − 5 = 9x + 8 → 4x = 9x + 8 − 5 → … → x = −5/3First change subtracts 5 instead of adding 5; last division is wrong tooAdd 5: 4x = 9x + 13; −5x = 13; x = −13/5; both original sides are −77/5
Common mistake

Do not “repair” a valid step merely because it is different from yours. Conversely, a correct-looking final fraction does not validate an earlier step. Check the first failed operation and then test the repaired answer in the original.

Quiz

Quick check

What is the first error in 4x + 6 = 10 → 4x = 16 → x = 4?

Quick check

Which equals −2(3 − 4x)?

Quick check

Dividing 3x + 1 = −12 by 3 gives which equation?

Quick check

Which step is valid?

Quick check

What is the correct solution of 4x − 5 = 9x + 8?

Practice Problems

Practice Problems
  1. For 6x + 9 = 66 → x + 9 = 11, identify the first invalid step, repair it, and check.
  2. Explain why every line of 14y + 24 = 36 → 7y + 12 = 18 → 7y = 6 → y = 6/7 is valid.
  3. A learner expands −3(u + 2) as −3u + 6. Explain the sign mistake and expand correctly.
  4. Correct 2(a + 5) = 18 → 2a = 13. Show the missing multiplication.
  5. For 5b = 2b + 9 → 7b = 9, explain the correct operation and solve.
  6. Create an incorrect two-line equation solution involving division of a sum. Then explain and repair your own error.
  7. Choose one repaired solution above and show both the original LHS and RHS as exact fractions.

Key Takeaways

Key Takeaways

• Find the first step that fails to preserve equality. • A coefficient is undone by division, not subtraction. • Divide or multiply every term in a grouped side. • Keep letters attached to unknown terms and track negative signs. • Recognise valid alternative methods and verify every repaired answer.

Previous · Lesson 8

Number Tricks and Equation Chains

Next · Lesson 10

A Pinch of History