Historical Problem Studies: Algebra

**By: Tao Steven Zheng (鄭濤)**

## **Introduction**

The word **algebra** comes from the Arabic term *al-jabr* (الجبر) which roughly translates to " the reunion of broken parts". The word originated from the title *al-Kitāb al-Mukhtaṣar fī Ḥisāb al-Jabr wal-Muqābalah* (ٱلْكِتَاب ٱلْمُخْتَصَر فِي حِسَاب ٱلْجَبْر وَٱلْمُقَابَلَة, *The Compendious Book on Calculation by Completion and Balancing*), a mathematical treatise written by the Persian mathematician and astronomer Muhammed ibn Musa al-Khwarizmi (c. 780 – c. 850 AD) in early 9th century. However, algebraic problems have certainly existed before al-Khwarizmi and was studied extensively by several ancient civilizations, such as the ancient Egyptians, Babylonians, Chinese, Greeks, and Indians. These ancient cultures came to realize the necessity of being able to solve for unknown quantities, and subsequently developed several ingenious methods for solving such problems. For most of human history, people have expressed their equations and solution methods rhetorically (i.e. in words), and had perhaps developed computation tools or shorthand when performing the necessary computations. In modern times, algebraic equations are expressed using a combination of letters and numbers to represent variables and coefficients respectively. The practice of solving algebraic equations requires the mastery of proper manipulation of the variables and coefficients. In this section, we will examine a number of problems taken from ancient mathematical documents to illustrate how ancient peoples would have solved them. The ancient methods will then be compared to the modern method to reveal any similarities or differences in thinking.

## **Ancient Egyptian Algebra**

### Linear Equations

There are many problems recorded on Egyptian papyri that involve linear equations. Among the earliest and most basic linear equations were called **Aha problems**. Problem 19 of the *Moscow Mathematical Papyrus* (c. 1850 BC), a good example of an Aha problem, asks:

*A number taken* 1 *and* 1/2 *times and added to* 4 *to make* 10*. What is the number?*

**Solution**

The scribe gives a solution:

*Calculate the excess of this* 10 *over* 4*. The result is* 6*. You operate on* 1 and 1/2 *to find* 1*. The result is* 2/3*. You take* 2/3 *of this* 6*. The result is* 4*. Behold,* 4 *says it. You will find that this is correct.*

In modern notation, the equation is

x

1
2

x
# 4

{\displaystyle x + \frac{1}{2}x + 4 = 10. }

The ancient Egyptian scribe instructs the reader to subtract 4 from 10

x

1
2

# x

{\displaystyle x + \frac{1}{2}x = 6. }

After subtracting 4, the scribe noted that the reciprocal of

1

1
2

{\displaystyle 1 \frac{1}{2}}

is equal to

2
3

{\displaystyle \frac{2}{3}}

. The scribe then multiplies 6 by

2
3

{\displaystyle \frac{2}{3}}

to obtain the answer:

2
3

×
# 6

{\displaystyle \frac{2}{3} \times 6 = 4. }

The above solution is similar to the method taught in school mathematics today; however, the ancient Egyptians often utilized a different method when solving linear equations. This technique is called the **method of false position**, or *regula falsi* in Latin. The method of false position begins by assuming a convenient but probably incorrect answer. The incorrect answer is then adjusted by the correct proportion to arrive at the correct answer. A good example of this method can be found in Problem 26 of the *Rhind Mathematical Papyrus* (c. 1650 BC):

*Find a quantity such that when it is added to* 1/4 *of itself the result is a* 15*.*

**Solution**

The scribe solves the problem as follows:

*Assume* [the answer is] 4*. Then* 1 *and* 1/4 *of* 4 *is* 5*. Multiply* [the number to] 5 *so as to get* 15*. The answer is* 3*. Multiply* 3 *by* 4*. The answer is* 12*.*

In modern notation, the equation is

x

1
4

# x

{\displaystyle x + \frac{1}{4}x = 15. }

The scribe makes an initial guess

# x
4

{\displaystyle x=4}

based on the denominator of the fraction 1/4, and then proceeds to calculate the left hand side of the equation:

4

1
4

×
# 4

{\displaystyle 4 + \frac{1}{4} \times 4 = 5. }

The scribe would notice that the initial guess gives a result that is one-third of 15. He then concludes that the correct answer must be triple the initial guess:

# x
3
×
# 4

{\displaystyle x = 3 \times 4 = 12. }

Therefore, the desired number is 12. The *Rhind Mathematical Papyrus* has several problems of this type, all solved using the method of false-position. There is no discussion of how or when this method was discovered or any explanation of why it works. However, it is clear that the ancient Egyptian scribes understood the concept of proportionality of two quantities in a linear relationship of the form

# y
a
x

{\displaystyle y = ax}

, where

x

{\displaystyle x}

 is the variable to be solved,

y

{\displaystyle y}

is the given quantity, and

a

{\displaystyle a}

is the proportionality constant.

### Quadratic Equations

How did the ancient Egyptians solve quadratic equations? The near absence of quadratic problems from surviving Egyptian mathematics texts makes this question difficult to answer. Of all the the archaeological evidence available, the earliest extant quadratic equation problems from ancient Egypt can only be found on the surviving fragments of the *Berlin Papyrus 6619* (c. 1800 BC). This document contains only two problems, of which Problem 1 states:

*You are told the area of a square of* 100 *square cubits is equal to that of two smaller squares. The side of one square is* 1/2 + 1/4 *of the other. What are the sides of the two unknown squares?*

**Solution**

Let

x

{\displaystyle x}

 be the side length of the small square and

y

{\displaystyle y}

 be the side length of the large square. According to the problem, we get a system of simultaneous equations:

{

x
2

y
2

=
100

# x

(

1
2

1
4

)

y

.

{\displaystyle \begin{align} \begin{cases} x^2 + y^2 = 100 \\ x = \left(\frac{1}{2} + \frac{1}{4} \right)y \end{cases} \end{align}. }

Note that the second equation is equivalent to

# x

3
4

y

{\displaystyle x = \frac{3}{4}y }

but is expressed as

# x

(

1
2

1
4

)

y

{\displaystyle x = \left(\frac{1}{2} + \frac{1}{4} \right) y }

 because the ancient Egyptians expressed their fractions as the sum of unit fractions. Interestingly, the ancient Egyptians applied the method of false-position on quadratic equations as well! The scribe who wrote the *Berlin Mathematical Papyrus*, begins by assuming that the sides of the two squares are 1 and 3/4. He then calculated the sum of the areas of these two squares to be

1
2

(

3
4

)

2

=
1

9
16

.

{\displaystyle 1^2 + \left(\frac{3}{4} \right)^2 = 1 \frac{9}{16}.}

The scribe realized that he could not compare each area directly too 100 (the total area), so he must compare their sides. Subsequently, the scribe took the square root of

1

9
16

{\displaystyle 1\frac{9}{16}}

, which is

1

1
4

{\displaystyle 1\frac{1}{4}}

. The scribe than compared this to 10 (the square root of 100). Since 10 is 8 times larger than

1

1
4

{\displaystyle 1\frac{1}{4}}

, the scribe concluded that the sides of the two other squares must be 8 times the original guesses.

{

8
×
1

=
8

8
×

3
4

=
6

{\displaystyle \begin{cases}
8 \times 1 &= 8 \\
8 \times \frac{3}{4} &= 6
\end{cases}}

Thus, the side of the large square is 8 and the side of the small square is 6 cubits.

## **Ancient Babylonian Algebra**

The ancient Babylonians were solving algebraic problems around the same time as the ancient Egyptians. The Babylonians wrote their problems on clay tablets which were a lot more durable than Egyptian papyri. Curiously, many mathematical clay tablets were in fact homework problems passed down from teacher to student. The vast collection of these unearthed documents provide archaeologists and historians excellent insight into the mathematical mind of ancient Babylon. We now know that the ancient Babylonians solved algebraic equations using an algorithmic approach, and their level of sophistication was unrivalled for their time. Babylonian mathematicians (and their pupils) were capable of solving linear and quadratic equations, and certain special cases for cubic and quartic equations.

### Linear Equations

Like the ancient Egyptians, the Babylonians also solved linear equations using the method of false position. For example, a problem from tablet YBC 4652 states:

*I found a stone, but did not weigh it; after I added one-seventh and then one-eleventh [of the total], it weighed* 1 *mina . What was the original weight of the stone?*

Note: 1 *mina* = 60 *gin*

**Solution**

Begin by converting 1 *mina* to 60 *gin*. According to the problem, this equation is:

(

x

x
7

)

1
11

(

x

x
7

)

=

{\displaystyle \left(x + \frac{x}{7}\right) + \frac{1}{11} \left(x + \frac{x}{7}\right) = 60.}

The author of this clay tablet does not reveal any solution method, giving only the answer. If the problem was solved by the method of false position, he would have made the initial guess

x

x
7

=
11

{\displaystyle x + \frac{x}{7}= 11}

, effectively transforming the problem into

y

1
11

# y

{\displaystyle y + \frac{1}{11}y = 60.}

Evaluating the left hand side of the equation gives

11

1
11

×
# 11

{\displaystyle 11 + \frac{1}{11}\times 11 = 12.}

Since

60
÷
# 12
5

{\displaystyle 60 \div 12 = 5}

, the guess must be multiplied by 5, which gives

x

x
7

=
11
×
# 5

{\displaystyle x + \frac{x}{7} = 11 \times 5 = 55.}

Next solve the equation

x

x
7

=
55
,

{\displaystyle x + \frac{x}{7} = 55,}

again, using the method of false position . The initial guess this time would be

# x
7

{\displaystyle x=7}

; thus,

7

7
7

=
8

{\displaystyle 7 + \frac{7}{7} = 8}

The last step would be to multiply the guess

# x
7

{\displaystyle x=7}

by

55
8

{\displaystyle \frac{55}{8}}

to get

7
×

55
8

=

385
8

=
48

1
8

.

{\displaystyle 7 \times \frac{55}{8} = \frac{385}{8} = 48\frac{1}{8}.
}

### Quadratic Equations

One famous ancient Babylonian problem on quadratic equations is found on tablet BM 13901:

*The area of a square less its side is* 14, 30 [870]*. What is the side of this square?*

**Solution**

The Babylonian sexagesimal (base 60) number 14, 30 is equivalent to 870 in decimal (base 10). Thus, we would model this problem as

x
2

−
# x
870

{\displaystyle x^2 - x = 870}

in modern mathematical notation. The Babylonians often give their solution as a set of instructions for finding the unknown:

*Take half of* 1*, which is* 30 [*sixtieths*]*. Multiply it by itself to get* 15 [*sixtieths*]*. Add this to* 14, 30 *to get* 14, 30; 15 [870:25]*. Take the square root to get* 29; 30 [29:5]*. Now add* 30 [*sixtieths*] *to this number to get* 30*, which is the side of the square.*

More generally, the above description gives the formula

# x

(

b
2

)

2

c

b
2

{\displaystyle x=\sqrt{{\left(\frac{b}{2}\right)}^{2} + c} + \frac{b}{2} }

for finding the positive solution of the quadratic equation

x
2

−
b
# x
c

{\displaystyle x^2 - bx = c}

. This solution method is based on the geometrically-inspired technique of completing the square. Simply substituting the coefficients of the equation yields:

x

=

(

b
2

)

2

c

b
2

=

(

1
2

)

2

870

1
2

=
30

{\displaystyle \begin{align}
x &= \sqrt{{\left(\frac{b}{2}\right)}^{2} + c} + \frac{b}{2} \\
&= \sqrt{{\left(\frac{1}{2}\right)}^{2} + 870} + \frac{1}{2} \\
&= 30
\end{align} }

The ancient Babylonians, being accustomed to working in base 60 expressed their fractions in one-sixtieth. This explains why fractions like 1/2 and 900/3600 are expressed as 30/60 and 15/60 respectively.

## **Ancient Chinese Algebra**

Prior to the Enlightenment in Europe, the Chinese civilization had developed the most advanced methods for solving algebraic equations. Chinese mathematicians arranged counting rods to represent the coefficients of simultaneous linear equations and the computation process in each intermediate step. They also developed sophisticated numerical algorithms for solving equations of any degree. A system for representing multivariate polynomials using counting rods was also developed.

### Linear Equations

Much of what we know about the ancient Chinese study of linear equations can be found in the mathematics text called the *Jiuzhang Suanshu* (九章算術, *Nine Chapters on the Mathematical Art*). We will refer to this text as the *Nine Chapters*. Chapter 3 of the *Jiuzhang Suanshu*, titled *Cui fen* (衰分, "proportional distribution"), is concerned with problems dividing some quantity among a given set of progressive rates (列衰 *lie cui*). Chapter 3 Problem 4 of *Nine Chapters* asks:

*Suppose there is a woman who is skilled at weaving. Every day she doubles her product. In* 5 *days she weaves* 5 *chi. How much does she weave on each day?*

Note: 1 *chi* = 10 *cun*

**Solution**

The total length of the woven fabric after 5 days measures 5 *chi*, or 50 *cun*. Since the weaver doubles her product each successive day, the progressive rates of distribution (列衰 *lie cui*) are

1
,
2
,
4
,
8
,
16

{\displaystyle {1, 2, 4, 8, 16} }

. According to this problem, we get the following linear equation:

x
2
x
4
x
8
x
16
# x

{\displaystyle x + 2x + 4x + 8x + 16x = 50. }

After collecting all the linear terms on the left hand side, we solve the equation:

31
# x
50

{\displaystyle
31x = 50 }

# x

50
31

=
1

19
31

{\displaystyle
x = \frac{50}{31} = 1\frac{19}{31}
 }

Lastly, multiply the value

# x
1

19
31

{\displaystyle x = 1\frac{19}{31} }

 to each rate of distribution to obtain the weaver's product on each day:

1

19
31

c
u
n
,
3

7
31

c
u
n
,
6

14
31

c
u
n
,
12

28
31

c
u
n
,
25

25
31

c
u
n
.

{\displaystyle 1\frac{19}{31} \; cun, 3\frac{7}{31} \; cun, 6\frac{14}{31} \; cun, 12\frac{28}{31} \; cun, 25\frac{25}{31} \; cun. }

In the original answer found in the *Nine Chapters*, the last two lengths are expressed as a combination of *chi* and *cun*:

1

c
h
i

2

28
31

c
u
n

{\displaystyle 1 \;chi \; 2\frac{28}{31} \; cun }

and

2

c
h
i

5

25
31

c
u
n

{\displaystyle 2 \; chi \; 5\frac{25}{31} \; cun }

.

Answer:

1

19
31

c
u
n
,
3

7
31

c
u
n
,
6

14
31

c
u
n
,
12

28
31

c
u
n
,
25

25
31

c
u
n
.

{\displaystyle 1\frac{19}{31} \; cun, 3\frac{7}{31} \; cun, 6\frac{14}{31} \; cun, 12\frac{28}{31} \; cun, 25\frac{25}{31} \; cun. }

The ancient Chinese were known to have solved systems of simultaneous linear equations in two unknowns using an algorithm called the *ying bu zu shu* (盈不足術, "method of excess and deficiency"). This algorithm is in fact a type of linear interpolation more commonly called the **method double false position** today. This method can be traced to Chapter 7 of the *Nine Chapters*, entitled *ying bu zu* (盈不足, "excess and deficiency"), from which the algorithm is named. Each problem begins with two "false assumptions": one being an overestimate, or excess; the other being an underestimate, or deficiency. Both values are then substituted into a set formulas that gives the "right fit".

If one lets

(

a
1

,

b
1

)

{\displaystyle (a\_1, b\_1) }

 represent the excess,

(

a
2

,

b
2

)

{\displaystyle (a\_2, b\_2) }

 represent the deficiency,

x

{\displaystyle x}

 the number of people, and

y

{\displaystyle y}

 the price of the item, these problems can be modeled as a system of simultaneous linear equations in two unknowns:

{

a
1

x
−
# y

b
1

a
2

x
−
# y
−

b
2

{\displaystyle \begin{cases} a\_1 x - y = b\_1 \\ a\_2 x - y = - b\_2 \end{cases} }

According to the *ying bu zu shu* (method of excess and deficiency) the formulas for determining the number of people and the price of the item are

# x

b
1

b
2

a
1

−

a
2

{\displaystyle x = \frac{b\_1 + b\_2}{|a\_1 - a\_2|} }

and

# y

a
1

b
2

a
2

b
1

a
1

−

a
2

{\displaystyle y = \frac{a\_1 b\_2 + a\_2 b\_1}{|a\_1 - a\_2|} }

respectively.

The *ying bu zu* method gives the exact solution for linear equations; subsequently, when the method is used for solving systems of linear equations of two unknowns, it closely resembles the determinant calculations in Cramer's Rule for a 2x2 linear system. Chapter 7 Problem 1 of the *Nine Chapters* asks:

*Suppose there is an item purchased jointly. If each person pays* 8 [*qian*]*, there is an excess of* 3 [*qian*]*; if each person pays* 7 [*qian*]*, there is a deficiency of* 4 [*qian*]*. Question: what is the price of the item, and the number of people?*

**Solution**

For this problem, the excess is

(

a
1

,

b
1

# )
(
8
,
3
)

{\displaystyle (a\_1, b\_1) = (8,3) }

 and the deficiency is

(

a
2

,

b
2

# )
(
7
,
4
)

{\displaystyle (a\_2, b\_2) = (7,4) }

. Substituting each value into the *ying bu zu* formulas yields:

# x

3
4

8
−
7

=
7

{\displaystyle x = \frac{3 + 4}{|8 - 7|} = 7 }

# y

8
×
4
7
×
3

8
−
7

=
53

{\displaystyle y = \frac{8 \times 4 + 7 \times 3}{|8 - 7|} = 53 }

Answer: 7 people; price of the item is 53 [*qian*].

   The Chinese were the first to discover a systematic method for solving simultaneous linear equations using arithmetic operations and called it the *fang cheng shu* (方程術, "method of rectangular arrays"). Chapter 8 of the *Nine Chapters on the Mathematical Art*, entitled *fang cheng* (方程, "rectangular arrays"), provides the earliest description this method. In practice, the *fang cheng shu* (method of rectangular arrays) begins by instructing the mathematician to place counting rods in a rectangular array to represent the coefficients of linear equations, with each equation confined to a column. The mathematician then manipulates the rods by carrying out the *fang cheng* algorithm in order to eliminate more and more entries until what remains are the entries along the main diagonal (from top right to bottom left) and the entries in the bottom row. The solution set to the system of simultaneous linear equations can then be determined from the remaining entries. This practice is identical to the modern use of matrices and the elimination method when solving problems of this nature. In the West, this method is attributed to Carl Friedrich Gauss (1777 - 1855 AD), but was likely discovered roughly a century earlier by mathematicians such as Isaac Newton (1642 - 1727 AD).

Chapter 8 Problem 1 of the *Nine Chapters* asks:

*Suppose there are* 3 *bundles of top-grade millet,* 2 *bundles of medium-grade millet, and* 1 *bundle of low-grade millet, totaling* 39 *dou;* 2 *bundles of top-grade millet,* 3 *bundles of medium-grade millet, and* 1 *bundle of low-grade millet, totaling* 34 *dou;* 1 *bundles of top-grade millet,* 2 *bundles of medium-grade millet, and* 3 *bundle of low-grade millet, totaling* 26 *dou. Question: how much is the total* [*in dou*] *of* 1 *bundle of top-grade, medium-grade, and low-grade millet?*

**Solution**

This problem is characterized by the system of simultaneous linear equations in three variables:

{

3
x
2
y
# z
39

2
x
3
y
# z
34

x
2
y
3
# z
26

{\displaystyle \begin{cases} 3x + 2y + z = 39 \\
2x + 3y + z = 34 \\
x + 2y + 3z = 26
\end{cases} }

where

x

{\displaystyle x}

 represents 1 bundle of high-grade millet,

y

{\displaystyle y}

 represents 1 bundle of medium-grade millet,

z

{\displaystyle z}

 represents 1 bundle of low-grade millet.

The ancient Chinese mathematician would arrange counting rods to represent the coefficients of the equations in vertical columns from right to left. This is different from the horizontal arrangement in modern matrix notation. The solution below showcases a side-by-side comparison of both notations.

[

1

2

3

2

3

2

3

1

1

26

34

39

]

{\displaystyle \begin{bmatrix}
 1 & 2 & 3 \\
 2 & 3 & 2 \\
 3 & 1 & 1 \\
 26 & 34 & 39
 \end{bmatrix}
}

[

3

2

1

39

2

3

1

34

1

2

3

26

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 2 & 3 & 1 & 34 \\
 1 & 2 & 3 & 26\\
 \end{bmatrix}
}

The first step is to eliminate the top entry in the middle column. This is achieved by scaling the entire middle column with the top entry in the right column, which is 3, and then subtract the middle column by the right column twice. The eliminated entry is then left empty.

[

1

6

3

2

9

2

3

3

1

26

102

39

]

{\displaystyle \begin{bmatrix}
 1 & 6 & 3 \\
 2 & 9 & 2 \\
 3 & 3 & 1 \\
 26 & 102 & 39
 \end{bmatrix}
}

[

1

3

2

5

2

3

1

1

26

24

39

]

{\displaystyle \begin{bmatrix}
 1 & & 3 \\
 2 & 5 & 2 \\
 3 & 1 & 1 \\
 26 & 24 & 39
 \end{bmatrix}
}

[

3

2

1

39

6

9

3

102

1

2

3

26

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 6 & 9 & 3 & 102 \\
 1 & 2 & 3 & 26\\
 \end{bmatrix}
}

[

3

2

1

39

0

5

1

24

1

2

3

26

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 0 & 5 & 1 & 24 \\
 1 & 2 & 3 & 26\\
 \end{bmatrix}
}

The second step is to eliminate the top entry of the left column. This is achieved by scaling the entire left column with the top entry in the right column, which is 3, and then subtract the left column by the right column.

[

3

3

6

5

2

9

1

1

78

24

39

]

{\displaystyle \begin{bmatrix}
 3 & & 3 \\
 6 & 5 & 2 \\
 9 & 1 & 1 \\
 78 & 24 & 39
 \end{bmatrix}
}

[

3

4

5

2

8

1

1

39

24

39

]

{\displaystyle \begin{bmatrix}
 & & 3 \\
 4 & 5 & 2 \\
 8 & 1 & 1 \\
 39 & 24 & 39
 \end{bmatrix}
}

[

3

2

1

39

0

5

1

24

3

6

9

78

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 0 & 5 & 1 & 24 \\
 3 & 6 & 9 & 78\\
 \end{bmatrix}
}

[

3

2

1

39

0

5

1

24

0

4

8

39

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 0 & 5 & 1 & 24 \\
 0 & 4 & 8 & 39 \\
 \end{bmatrix}
}

The third step is to eliminate the second entry of the left column. This is achieved by scaling the entire left column with the second entry in the middle column, which is 5, and then subtract the left column by the right column four times. Finally reduce all the entries of the left column by 9.

[

3

20

5

2

40

1

1

195

24

39

]

{\displaystyle \begin{bmatrix}
 & & 3 \\
 20 & 5 & 2 \\
 40 & 1 & 1 \\
 195 & 24 & 39
 \end{bmatrix}
}

[

3

5

2

36

1

1

99

24

39

]

{\displaystyle \begin{bmatrix}
 & & 3 \\
 & 5 & 2 \\
 36 & 1 & 1 \\
 99 & 24 & 39
 \end{bmatrix}
}

[

3

5

2

4

1

1

11

24

39

]

{\displaystyle \begin{bmatrix}
 & & 3 \\
 & 5 & 2 \\
 4 & 1 & 1 \\
 11 & 24 & 39
 \end{bmatrix}
}

[

3

2

1

39

0

5

1

24

0

4

8

39

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 0 & 5 & 1 & 24 \\
 0 & 4 & 8 & 39 \\
 \end{bmatrix}
}

[

3

2

1

39

0

5

1

24

0

0

36

99

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 0 & 5 & 1 & 24 \\
 0 & 0 & 36 & 99 \\
 \end{bmatrix}
}

[

3

2

1

39

0

5

1

24

0

0

1

11
4

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 0 & 5 & 1 & 24 \\
 0 & 0 & 1 & \frac{11}{4} \\
 \end{bmatrix}
}

From here, one uncovers the third unknown quantity

# z

11
4

=
2

3
4

{\displaystyle z = \frac{11}{4} = 2\frac{3}{4} }

. Back-substitution of this value eliminates the third entry of the middle and right columns. Solving for the second unknown quantity gives

# y

17
4

=
4

1
4

{\displaystyle y = \frac{17}{4} = 4\frac{1}{4} }

.

[

3

4

2

4

1

11

17

39

]

{\displaystyle \begin{bmatrix}
 & & 3 \\
 & 4 & 2 \\
 4 & & 1 \\
 11 & 17 & 39
 \end{bmatrix}
}

[

12

4

8

4

11

17

145

]

{\displaystyle \begin{bmatrix}
 & & 12 \\
 & 4 & 8 \\
 4 & & \\
 11 & 17 & 145
 \end{bmatrix}
}

[

3

2

1

39

0

5

0

85
4

0

0

1

11
4

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 1 & 39 \\
 0 & 5 & 0 & \frac{85}{4} \\
 0 & 0 & 1 & \frac{11}{4} \\
 \end{bmatrix}
}

[

3

2

0

145
4

0

1

0

17
4

0

0

1

11
4

]

{\displaystyle \begin{bmatrix}
 3 & 2 & 0 & \frac{145}{4} \\
 0 & 1 & 0 & \frac{17}{4} \\
 0 & 0 & 1 & \frac{11}{4} \\
 \end{bmatrix}
}

Another back-substitution of

# y
4

1
4

{\displaystyle y = 4\frac{1}{4} }

 eliminates the second entry of the right column. Solving for the first unknown quantity gives

# x

37
4

=
9

1
4

{\displaystyle x = \frac{37}{4} = 9\frac{1}{4} }

.

[

12

4

4

11

17

111

]

{\displaystyle \begin{bmatrix}
 & & 12 \\
 & 4 & \\
 4 & & \\
 11 & 17 & 111
 \end{bmatrix}
}

[

4

4

4

11

17

37

]

{\displaystyle \begin{bmatrix}
 & & 4 \\
 & 4 & \\
 4 & & \\
11 & 17 & 37
 \end{bmatrix}
}

[

3

0

0

111
4

0

1

0

17
4

0

0

1

11
4

]

{\displaystyle \begin{bmatrix}
 3 & 0 & 0 & \frac{111}{4} \\
 0 & 1 & 0 & \frac{17}{4} \\
 0 & 0 & 1 & \frac{11}{4} \\
 \end{bmatrix}
}

[

1

0

0

37
4

0

1

0

17
4

0

0

1

11
4

]

{\displaystyle \begin{bmatrix}
 1 & 0 & 0 & \frac{37}{4} \\
 0 & 1 & 0 & \frac{17}{4} \\
 0 & 0 & 1 & \frac{11}{4} \\
 \end{bmatrix}
}

Answer: 1 bundle of top-grade millet is

9

1
4

{\displaystyle 9 \frac{1}{4} }

 *dou*; 1 bundle of medium-grade millet is

4

1
4

{\displaystyle 4 \frac{1}{4} }

*dou*; 1 bundle of low-grade millet is

2

3
4

{\displaystyle 2 \frac{3}{4} }

*dou*.

## **Practice Problems**

**Problem 1:** Solve *Berlin papyrus 6619* Problem 1 using modern algebraic methods.

Solution