This MCQ module is based on: Mend the Mistake and History of Equations
TOPIC 22 OF 23
Mend the Mistake and History of Equations
🎓 Class 7
Mathematics
CBSE
Theory
Ch 7 — Simple Equations
⏱ ~35 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Mend the Mistake and History of Equations
Targeting Class 7 level in General Mathematics, with Basic difficulty.
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7.3 Mind the Mistake, Mend the Mistake
Learning to spot and fix errors is as important as solving correctly. Below are six worked examples where a student made one or more slips. For each, your task is to (a) find the mistake, (b) describe it in words, and (c) redo the step correctly.
1
Student's work
4x + 6 = 10 4x + 10 = 6 4x = 16 − 10 x = 4
Mistake: the +6 was moved to the other side keeping the sign +6 and treating the 10 as -10. Correct step: subtract 6 from both sides: \(4x = 10 - 6 = 4\), so \(x = 1\).
2
Student's work
7 − 8z = 5 8z = 7 − 5 8z = 2 z = 2 / 8 = 1/4
Mistake: sign error. The coefficient of z is −8, but the student dropped the negative. Correct: \(7 - 8z = 5 \Rightarrow -8z = -2 \Rightarrow z = \tfrac{-2}{-8} = \tfrac{1}{4}\). In this case the final answer happens to be right, but the middle line \(8z = 2\) is wrong — it should be \(-8z = -2\).
3
Student's work
2v − 4 = 6 v − 4 = 3 v = 3 + 4 = 7
Mistake: the student divided only the term \(2v\) by 2, not the full equation. Correct: divide every term by 2, or first isolate \(2v\): \(2v = 10 \Rightarrow v = 5\).
4
Student's work
5z + 2 = 3z − 4 5z + 3z = −4 − 2 8z = −6 z = −6/8
Mistake: when moving \(3z\) across '=', the sign should change to \(-3z\), not \(+3z\). Correct: \(5z - 3z = -4 - 2 \Rightarrow 2z = -6 \Rightarrow z = -3\).
5
Student's work
15w − 4w = 26 7w + 12 = 26 7w = 26 − 12 7w = 14 w = 2
Mistake: the student wrote the original line as \(15w - 4w\) but later inserted "+12" without showing where it came from. The original problem may have been \(15w - 4w + 12 = 30\) or similar — the copying step is incorrect. Correct if equation is \(15w - 4w = 26\): combine like terms \(11w = 26 \Rightarrow w = \tfrac{26}{11}\).
6
Student's work
3x + 1 = −12 x + 1 = −12/3 x + 1 = −4 x = −5
Mistake: divided only \(3x\) by 3 without dividing the whole RHS by 3 and forgetting to also divide the constant +1. Correct: \(3x + 1 = -12 \Rightarrow 3x = -13 \Rightarrow x = -\tfrac{13}{3}\).
Common Slip Alert
Watch out for these three most-common slips:
- Sign-flip error: forgetting to change sign when moving a term across '='.
- Partial division: dividing only the variable term and not the constant.
- Wrong combine: combining unlike terms (a number with a variable).
Example 11 — Riyaz's Magic Trick
Riyaz asked Akash to perform the following steps:
- Think of a number.
- Subtract 3 from the number.
- Multiply the result by 4.
- Add 8 to the product.
- Reveal the final answer.
Akash's final answer was 24. Can Riyaz figure out the starting number? Yes — by translating the steps into an expression.
| Steps | Expression |
|---|---|
| Think of a number | \(x\) |
| Subtract 3 from the number | \(x - 3\) |
| Multiply the result by 4 | \(4(x - 3) = 4x - 12\) |
| Add 8 to the product | \(4x - 12 + 8 = 4x - 4\) |
Since Akash's final answer was 24, \(4x - 4 = 24 \Rightarrow 4x = 28 \Rightarrow x = 7\). So Akash thought of 7.
Quick rule to get the starting number from the final answer: add 4 to the final, then divide by 4. \( \dfrac{24 + 4}{4} = 7\).
Example 12 — Ramesh and Suresh's Marbles
Ramesh and Suresh have 60 marbles between them. Ramesh has 30 more marbles than Suresh. How many marbles does each boy have?
Let Ramesh have \(x\) and Suresh have \(y\). We have two unknowns but two conditions:
- \(x + y = 60\) (total marbles)
- \(x - y = 30\) (Ramesh has 30 more)
Using the second equation, \(x = y + 30\). Substituting into the first: \(y + 30 + y = 60 \Rightarrow 2y + 30 = 60 \Rightarrow 2y = 30 \Rightarrow y = 15\). So Suresh has 15 and Ramesh has 45.
Ramesh has 15 (same as Suresh) + 30 extra = 45. Total = 15 + 45 = 60.
Generating Equations
So far we have been handed an equation and asked to solve it. Now we can turn situations into equations. The skill of translating a word problem into symbols is the heart of algebra. A good habit: choose clear letter-names for unknowns, list every piece of given information, and write one equation for each relationship.
Here is a chain of transformations starting from \(y + 1 = 6\) and \(3y = 15\), where one is obtained from the previous by performing the same operation on both sides. Can you find the value of the unknown in each equation?
| Chain from \(y + 1 = 6\) | Same op on \(3y = 15\) |
|---|---|
| \(y + 1 = 6\) | \(3y = 15\) |
| Multiplying by \(-1\): \(-y - 1 = -6\) | Adding 6: \(3y + 6 = 21\) |
| Adding \(y\): \(-1 = -6 + y\) | Dividing by 3: \(y + 2 = 7\) |
| \(5 = y\) | \(y \times 2 = 7 \times 2 \Rightarrow 2y = 10\) |
7.4 A Pinch of History
Forming expressions using symbols and solving equations with unknowns is a very important component of mathematical exploration in algebra?. This story of algebra in India goes back thousands of years.
Indian Roots of Algebra
We have seen Brahmagupta's contributions to different areas of mathematics like integers and fractions. In Chapter 18 of his book Brāhmasphuṭasiddhānta (628 CE), he also explained how to add, subtract, and multiply unknown numbers using letters — the way we do today. This book was one of the earliest known works in algebra in history.
As renowned mathematician and Fields Medallist David Mumford puts it, "Brahmagupta is the key person in the creation of algebra as we know it."
In the 8th century, Indian mathematical ideas were translated into Arabic. They influenced a well-known mathematician named Al-Khwarizmi, who lived in present-day Iraq. Around 825 CE, he wrote a book called Hisab al-jabr wal-muqābala, which means "calculation by restoring and balancing." These ideas spread even further. By the 12th century, Al-Khwarizmi's book was translated into Latin and brought to Europe, where it became very popular. The word al-jabr from his book gave us the word algebra, which we also still use today.
Ancient Indian notation vs Modern Notation
| Modern Notation | Ancient Indian Notation |
|---|---|
| \(2x + 1\) | yā 2 rū 1 (in each term the indication of unknown/known comes first) |
| \(2x - 8\) | yā 2 rū 8̄ (a dot over the number indicates it is negative) |
| \(3x + 4 = 2x + 8\) | yā 3 rū 4 / yā 2 rū 8 (the two sides of an equation were presented one below the other) |
The symbols 'yā' (yāvattāvat) and 'rū' (rūpa) stood for the variable and the known number respectively. Brahmagupta also used colour-names to distinguish several different unknowns — kāla (black), nīla (blue), and so on. In place of a second variable, Indian mathematicians used particular letters (not the first letters of the colour names) that recurred systematically — much as we use \(x\) and \(y\) today.
Example 16 — Bijjaṇita (a problem from Bhāskarāchārya, 1150 CE)
One man has ₹300 rupees and 6 horses. Another man has 10 horses and ₹100 debt. If they are equally rich and the price of each horse is the same, tell me the price of one horse.
Solution: Let price of one horse = \(x\). According to the problem,
\(300 + 6x = 10x - 100\)
Subtract \(6x\) both sides: \(300 = 4x - 100\). Add 100 both sides: \(400 = 4x\). Divide by 4: \(x = 100\). Therefore the price of one horse is ₹100.
The power of a general formula
Such problems and solutions were well understood in ancient India. In fact, a very systematic way to solve problems with single unknowns was first proposed by Āryabhaṭa (499 CE) and Brahmagupta has outlined it in his Brāhmasphuṭasiddhānta (628 CE). Let us understand his method.
Let us look at a few equations of the following form:
\(5x + 4 = 3x + 8,\ 2x = 6 - 2x,\ 2x + 4 = 2\)
Can we come up with a formula to solve these equations? That is, for the first equation, can we perform some operations using 5, 4, 3 and 8 that will directly give us the solution? Using a similar method, can you solve the second equation using the numbers 5, ~6, 2 and 4?
To get a formula, we can generalise equations of this form by taking the four numbers as A, B, C and D. That is:
\(Ax + B = Cx + D\)
Brahmagupta's solution to equations of this form:
\(x = \dfrac{D - B}{A - C}\)
Using this approach, we can find the solution to the equation: \(650 + 4000 = 500m + 5050\), giving \(m = \dfrac{5050 - 4000}{650 - 500} = \dfrac{1050}{150} = 7\).
Why Generalise?
Ancient Indian mathematicians were thinkers of great insight. They introduced letter-symbols to solve mathematical problems. We have seen some examples in previous pages. By studying algebra, we learn to generalise patterns — in numbers, shapes, and situations. We address these generalisations using the language of algebra, which is both precise and powerful.
🧪 Activity: Become a Riyaz — Invent Your Own Number Trick
Materials: Paper, pencil, 2 friends
Predict: Can you design a four-step trick and discover the formula that recovers the starting number from the final answer?
- Think of four operations you will ask your friend to do (e.g., +5, ×2, −3, ÷2).
- Let \(x\) be the starting number. Write the expression after each step.
- Simplify the final expression. If your friend's final answer is \(F\), solve for \(x\) in terms of \(F\).
- Test the trick: ask your friend to think of a number, follow the steps, and tell you the final answer. Use your formula to name the starting number.
- Modify step 2 so the formula becomes \(x = F/2 + 1\). Challenge: design steps that give this formula.
Example: +5, ×2, −3, ÷2 gives \((2(x+5) - 3)/2 = x + \tfrac{7}{2}\). So starting number \(= F - 3.5\). If friend says "10.5", you say "7"!
Figure it Out (Section 7.3 & 7.4)
Q1. Identify and describe the mistake in: \(8x + 3 - 2 = 5 \Rightarrow 8x = 5 + 3 - 2 = 6 \Rightarrow x = \tfrac{6}{8}\).
Sign error: when moving \(+3\) and \(-2\) across the '=', both signs must flip. Correct: \(8x = 5 - 3 + 2 = 4 \Rightarrow x = \tfrac{1}{2}\).
Q2. Ramesh and Suresh have 76 marbles between them. Ramesh has 8 more marbles than Suresh. Find each boy's count.
\(x+y=76,\ x-y=8\). Adding: \(2x=84 \Rightarrow x=42\). So \(y=34\). Ramesh: 42, Suresh: 34.
Q3. Using Brahmagupta's formula, solve \(7x + 5 = 2x + 30\).
\(x = \tfrac{D-B}{A-C} = \tfrac{30 - 5}{7 - 2} = \tfrac{25}{5} = 5\).
Competency-Based Questions
Scenario: Three friends — Anya, Bilal and Charu — are collecting stamps. Anya has 12 more stamps than Charu. Bilal has twice as many as Charu. Together all three have 112 stamps.
Q1. Set up equations using \(c\) for Charu's stamps and find each person's count.
L3 ApplyAnswer: \((c+12) + 2c + c = 112 \Rightarrow 4c + 12 = 112 \Rightarrow 4c = 100 \Rightarrow c = 25\). So Charu 25, Anya 37, Bilal 50.
Q2. Aman solved the same problem and wrote \(c + 12 + 2c + c = 112 \Rightarrow 4c = 100 \Rightarrow c = 25\). Identify the missing step and explain whether his final answer is still correct.
L4 AnalyseAnswer: Aman skipped the line "\(4c + 12 = 112\)" — subtracting 12 from both sides. His answer is still correct, but skipping steps is risky: if he later had \(-12\) instead of \(+12\) he would likely make a sign slip. Always show every operation.
Q3. Evaluate: If instead Bilal had thrice as many stamps as Charu (instead of twice), does the problem still have a solution with positive integers? Justify.
L5 EvaluateAnswer: New equation: \((c+12) + 3c + c = 112 \Rightarrow 5c + 12 = 112 \Rightarrow c = 20\). So Charu 20, Anya 32, Bilal 60. Yes, all positive integers — valid.
Q4. Create a word problem whose algebraic model is \(2(x + 3) - 5 = x + 10\), and solve it.
L6 CreateSample: "I think of a number \(x\), add 3, double the result and subtract 5. My answer is the same as the sum of the original number and 10." Solve: \(2x + 6 - 5 = x + 10 \Rightarrow 2x + 1 = x + 10 \Rightarrow x = 9\).
Assertion–Reason Questions
A: Brahmagupta's formula \(x = \tfrac{D-B}{A-C}\) solves every equation of the form \(Ax + B = Cx + D\) provided \(A \neq C\).
R: When \(A = C\), the \(x\)-terms cancel and we are left with \(B = D\), which is either always true or always false.
R: When \(A = C\), the \(x\)-terms cancel and we are left with \(B = D\), which is either always true or always false.
Answer: (a) — R explains why the formula's denominator must be non-zero, i.e., why \(A \ne C\) is required.
A: The word "algebra" comes from the Arabic word al-jabr.
R: Al-Khwarizmi's 9th-century book introduced the word into Latin translations read across Europe.
R: Al-Khwarizmi's 9th-century book introduced the word into Latin translations read across Europe.
Answer: (a) — Both statements are factually correct, and R provides the historical mechanism for A.
A: If a student moves \(+6\) from one side of \(=\) to the other and writes it as \(+6\) again, the equation will always be wrong.
R: Moving a term across '=' requires changing its sign.
R: Moving a term across '=' requires changing its sign.
Answer: (a) — Both true; R is the rule that explains A.
Frequently Asked Questions
What is the 'Mend the Mistake' activity about?
It shows incorrectly solved equations and asks students to spot the error, explain it, and present the corrected solution. This sharpens attention to each algebraic step.
Who invented algebra?
The word 'algebra' comes from the Arabic 'al-jabr' used by Al-Khwarizmi around 825 CE. Indian mathematicians like Brahmagupta (7th century) and Bhaskara II (12th century) also contributed significantly to algebraic methods.
Why is the history of equations important?
Knowing the historical roots shows that algebraic ideas developed across cultures and that the methods we use today are refinements of techniques invented over centuries.
How can you avoid sign errors in transposition?
Write each step completely, circle the term being moved, flip its sign only, and double-check by substituting your answer into the original equation at the end.
What is a systematic check for a solved equation?
Replace the variable with the solution in the original equation and compute both sides separately. If they match, the solution is correct.
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