Finding the Unknown · Lesson 10 of 13
A Pinch of History
“Connect algebra’s changing notation with the same enduring idea of equality.”
• Describe the seed metaphor behind bijaganita. • Read the role of older symbols for unknowns and known quantities. • Solve the horse-price problem using equality. • Derive a general solution rule and state its restriction.
A small unknown can organise a whole problem
In Indian mathematical writing, bijaganita refers to algebra. Bija means seed: a small starting idea can grow into a method for many problems. An unknown letter plays a similar role. It lets us state a relationship before we know the number, then use that relationship to discover it. Algebra also explains why numerical patterns hold, rather than only checking examples.
For instance, any odd whole number can be written as twice a whole number plus 1. Two odd numbers can be represented by 2a + 1 and 2b + 1. Their sum is 2a + 2b + 2 = 2(a + b + 1), which is twice a whole number and therefore even. The letters explain the pattern for every pair, instead of relying on a few sums.
Ideas travelled; notation changed
Systematic equation methods have a long history. Aryabhata’s work is associated with 499 CE. Brahmagupta described arithmetic with unknown quantities in the Brāhmasphuṭasiddhānta of 628 CE. Indian mathematical ideas were translated into Arabic in the eighth century. Around 825 CE, Al-Khwarizmi wrote Hisab al-jabr wal-muqabala, a work whose title refers to calculation by restoring and balancing. Its ideas reached European readers through Latin translations in the twelfth century. The term al-jabr gave rise to the word algebra. These connections show how mathematical ideas develop and travel across languages.
Older Indian notation used abbreviated words. Yā, from yāvat-tāvat, could represent an unknown; rū, from rūpa, introduced a known numerical quantity. Colour names could distinguish different unknowns: kā referred to black and nī to blue. A dot above a number indicated a negative term in the notation illustrated here. We use modern letters and minus signs, but the underlying relationships remain the same.
| Modern expression | How the older notation records it |
|---|---|
| 2x + 1 | yā 2 rū 1: two unknowns and a known 1 |
| 2x − 8 | yā 2 rū 8, with a dot above 8 to indicate the negative term |
| 3x + 4 = 2x + 8 | Two aligned lines: yā 3 rū 4 above yā 2 rū 8; the lines represent equal quantities |
Problem
Explain what yā 2 rū 1 expresses in modern notation.
- 1.The abbreviation yā names an unknown quantity; the following 2 says there are two copies.
- 2.The abbreviation rū introduces the known quantity 1.
- 3.Using x for the unknown, the expression is 2x + 1. Changing the letter would change the notation, not the relationship.
A horse-price problem
An old problem associated with Bhaskara’s work, around 1150 CE, describes two people whose wealth becomes equal after the value of their horses is included. Give one horse’s price a letter, because the horses are taken to have the same value. Then each person’s wealth is an expression involving that one price.
Problem
One person has ₹300 and six horses. Another has ten horses but owes ₹100. Their total wealth is equal. Find the price of a horse.
- 1.Let x be the price of one horse in rupees. The first wealth is 300 + 6x. The second is 10x − 100, because debt reduces wealth.
- 2.Write 300 + 6x = 10x − 100. Subtract 6x: 300 = 4x − 100.
- 3.Add 100: 400 = 4x. Divide by 4: x = ₹100 per horse.
- 4.Check: first wealth = 300 + 600 = ₹900; second wealth = 1000 − 100 = ₹900.
Derive one rule for many equations
The same operations solve an entire family of equations. The chapter connects the following rule with Brahmagupta’s method. Let A and C be known coefficients of x, and B and D be known constant terms. In Ax + B = Cx + D, subtract Cx from both sides, then subtract B. This gives (A − C)x = D − B. If A − C is non-zero, divide by it to isolate x.
If A = C, this formula cannot be used because its denominator would be zero. Instead, cancel the equal unknown terms and inspect the remaining constants. Different constants give no solution, as in x + 4 = x + 5. If the constants are also equal, both sides are the same expression and every allowed value makes them equal. Return to the equation instead of dividing by zero.
Problem
Solve 5x + 4 = 3x + 8 using the rule and check it.
- 1.Identify A = 5, B = 4, C = 3, D = 8. Since 5 − 3 = 2 is non-zero, the rule applies.
- 2.x = (8 − 4)/(5 − 3) = 4/2 = 2.
- 3.Check: 5 × 2 + 4 = 14 and 3 × 2 + 8 = 14. The rule abbreviates the same subtraction and division steps already learned.
Problem
Solve 3x − 6 = 2x + 4, then 2x + 3 = 4x + 5.
- 1.For 3x − 6 = 2x + 4, B = −6. Thus x = [4 − (−6)]/(3 − 2) = 10/1 = 10.
- 2.Check: 30 − 6 = 20 + 4 = 24.
- 3.For 2x + 3 = 4x + 5, x = (5 − 3)/(2 − 4) = 2/(−2) = −1.
- 4.Check: 2(−1) + 3 = 1 and 4(−1) + 5 = 1.
In the general rule, B is the complete signed constant. For 3x − 6, B = −6, not 6. Never use the formula when A = C; inspect the simplified equality instead.
Quiz
What does the seed metaphor suggest about algebra?
What does rū introduce in the older notation described here?
How is a negative term indicated in the illustrated older notation?
For Ax + B = Cx + D, when may you divide by A − C?
If 300 + 6x = 10x − 100, what is x?
Practice Problems
- Explain bijaganita’s seed comparison in your own words using an equation from an earlier lesson.
- Translate “yā 3 rū 4” into a modern expression and explain both parts.
- Derive x = (D − B)/(A − C) starting from Ax + B = Cx + D. Name every operation.
- Use the derived rule to solve 7x + 2 = 4x + 17, then check.
- Apply the rule to the savings equation 650m + 4000 = 500m + 5050. State why the denominator is non-zero.
- Explain why x + 7 = x + 9 cannot be solved using a quotient with denominator zero.
- Use the form 2a + 1 to explain why adding two odd whole numbers gives an even whole number.
Key Takeaways
• An unknown acts as a starting idea for many relationships. • Algebra developed through mathematical work in several languages. • Notation changes while equality keeps its meaning. • The horse problem equates complete wealth, including debt. • The general solution rule comes from equal subtraction and requires unequal coefficients.