Mathematics · Expressions, equations and inequalities · Grade 7

The linear equation: the balance

The pans are equal: 3 sacks + 4 kg = 2 sacks + 13 kg. How many kilograms is one sack?

The situation

A market in al-Khwarizmi’s time. In one pan of the scale there are 3 identical sacks of rice and 4 kg of stones, in the other 2 such sacks and 13 kg of stones — the pans are level. How many kilograms is one sack of rice? We do not have to open the sacks and weigh them: we take identical things off both pans and the balance is not disturbed — in the end one sack and some stones remain. More than a thousand years ago al-Khwarizmi named this method “al-jabr and al-muqabala”.

Topic

The linear equation: the root, the balance model, al-Khwarizmi’s method, solving problems with equations. Al-Khwarizmi’s balance: unknown boxes and stones in two pans — the steps of al-jabr and al-muqabala

Key ideas

Equation and root
Al-jabr and al-muqabala
Problem → equation

Formulas

a x + b = c x + d

A linear equation in one unknown: $x$ is the unknown, $a, b, c, d$ are given numbers. The root is the value of $x$ that makes the equality true.

a x - c x = d - b \Rightarrow (a - c) x = d - b

Moving terms: a term goes from one side of the equality to the other with its sign changed (al-jabr and al-muqabala).

x = \frac{d - b}{a - c}, a \neq c

The root. If $a = c$ : for $b = d$ every $x$ is a root, for $b \neq d$ there is no root.

\frac{x + 1}{2} = 3 \Rightarrow x + 1 = 6

An equation with fractions: we multiply both sides by the denominator — the pans are doubled, the balance is kept.

Theory

The equation and its root

An equality with an unknown number in it is an equation: $3 x + 4 = 2 x + 13$ . The root of an equation is a number that makes the equality true when put in place of the unknown. For $x = 5$ the left side is $19$ and the right side $23$ — not equal; for $x = 9$ both are $31$ — so $9$ is a root. Solving an equation means finding all its roots (or showing that it has none). The balance model shows this: the pans are level only when the weight of a sack equals the root.

Al-Khwarizmi and his method

Abu Abdallah Muhammad ibn Musa al-Khwarizmi (born in Khorezm, c. 783–850) worked at the House of Wisdom (Bayt al-Hikma) in Baghdad. His book “Al-kitab al-mukhtasar fi hisab al-jabr wa-l-muqabala” teaches how to solve equations with two operations. Al-jabr (“restoring”) means removing a subtracted term (a debt) by adding it to both sides: from $x - 5 = 2 x - 12$ we get $x + 7 = 2 x$ . Al-muqabala (“balancing”) means removing identical terms from both sides: from $x + 7 = 2 x$ we get $7 = x$ . Al-Khwarizmi did not use negative numbers, so he divided equations into six types.

The linear equation and moving terms

An equation that, after simplifying, takes the form $a x + b = c x + d$ is called linear. The rule for solving: we move the terms with the unknown to one side and the numbers to the other — a moved term changes its sign (this is the same as subtracting the same term from both sides): $(a - c) x = d - b$ . If $a \neq c$ , there is one root: $x = \frac{d - b}{a - c}$ . If $a = c$ , the left side becomes $0 \cdot x$ : for $b = d$ we get $0 = 0$ — every $x$ is a root; for $b \neq d$ we get $0 = d - b$ , which is false — there is no root. On the balance: the number of sacks is the same, the stones differ — it will never balance.

Equations with brackets and fractions

If there are brackets, we first expand the brackets: from $3 (x + 2) = 5 x - 4$ we get $3 x + 6 = 5 x - 4$ , then as usual. If there are fractions, we multiply both sides by the denominator (by the common denominator if there are several): from $\frac{x + 1}{2} = 3$ we get $x + 1 = 6$ ; in $\frac{x - 1}{2} = \frac{x + 3}{4}$ the common denominator is $4$ : $2 (x - 1) = x + 3$ , hence $x = 5$ . If the unknown is in the denominator (a fractional linear equation, for example $\frac{3}{x - 1} = \frac{2}{x + 1}$ ), before multiplying we must make sure the denominator is not zero: we check the root found against the conditions $x \neq 1$ , $x \neq - 1$ — this continues in the “Algebraic fractions” experiment.

Solving problems with equations

Four steps. 1) Choose the unknown — usually the quantity asked for in the problem ( $x$ is the son’s age). 2) Express the other quantities in terms of $x$ (the father is $x + 24$ , in three years $x + 3$ and $x + 27$ ). 3) Set up the equation from the conditions of the problem: $x + 27 = 3 (x + 3)$ . 4) Solve it and check the answer against the problem: $x = 9$ — in three years $12$ and $36$ , $36 = 3 \cdot 12$ . The bar model shows steps 2 and 3: the father’s bar equals three of the son’s bars.

About the model

The balance is an explanatory model: the sacks are taken to have the same weight, the stones are exact, and there is no friction. A negative term is shown by a “ball” (a debt), but a balance cannot show an equation with a negative root — the rule, however, still works. The tilt of the pans is not like a real balance, it only tells which side is heavier. In the experiment the roots are shown as whole numbers or decimals (two places).

Did you know?

The word “algebra” comes from the word “al-jabr” in the title of al-Khwarizmi’s book “Al-kitab al-mukhtasar fi hisab al-jabr wa-l-muqabala”, written in Baghdad around 820.
The word “algorithm” comes from the Latin form of al-Khwarizmi’s name — “Algoritmi”: in the 12th century his book on arithmetic with Indian numerals was translated into Latin.

Inside the experiment

The topic is explained step by step with animation, from simple to advanced: from a real-life picture to the formula and the proof. Then you drag the points yourself to check. Inside are missions, a quiz and a guided tour.

More experiments in this section

Expressions: from numbers to letters — Grade 7
Monomials and polynomials — Grade 7
Special products (algebraic identities) — Grade 7
Algebraic fractions — Grade 7
Systems of equations — Grade 7
Inequalities: ray and interval — Grade 8
The quadratic equation: the parabola — Grade 8