Part II · Algebra Chapter 8 of 60
Equations: the art of balance
An equation is a pair of scales in balance with a box of unknown weight on it. We'll learn to take things off the pans without making the scales tip, and find out where the word “algebra” itself comes from.
Builds on: 7 · Letters for numbers
You will learn
- solve linear equations with brackets and fractions and check the answer by substitution
- understand which transformations keep the roots and which lose or add them
- translate a word problem into an equation and back
At the end of the last chapter a spectator told the magician: "I thought of a number, multiplied it by 3, added 7 and got 25". The magician can write what he heard in letters: $3x + 7 = 25$. But this is no longer an expression; it's a statement, and it's true not for every $x$ but for exactly one. That's the one we have to find.
You could guess. With $x = 5$ you get 22, too little; with $x = 7$ you get 28, too much; so the answer lies in between, and $x = 6$ works. Guessing worked because the numbers were small and the answer was a whole number. For the problem we'll reach in the middle of the chapter, $\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x$, it would be torture. We need a method, not luck. Such a method was invented more than a thousand years ago, and the best picture for it is a pair of balance scales.
Scales in balance
Put three identical boxes of unknown weight $x$ and two one-kilogram weights on the left pan, and fourteen weights on the right. The scales are level. That is the equation $3x + 2 = 14$: the left side sits on the left pan, the right side on the right pan, and the equals sign is the pivot.
An equation is an equality with an unknown, about which we ask: for which values of the unknown is it true? Each such value is called a root (or solution) of the equation. To solve an equation means to find all its roots or prove that it has none.
What can you do to the scales without upsetting the balance? Take the same thing off both pans, or put the same thing on both. And divide the contents of each pan into the same number of equal parts, keeping one part of each (or, the other way round, make both pans the same number of times heavier). Take two weights off each pan: three boxes remain on the left, twelve weights on the right, $3x = 12$. Split both pans into three parts: one box against four weights, $x = 4$. We've weighed the box without looking inside.
Scales can't hold a negative weight, so for subtraction we had to invent balloons. A balloon pulls its pan up with the force of one kilogram, and $4x - 3$ is four boxes with three balloons tied on. A weight and a balloon on the same pan cancel each other, like $+1$ and $-1$ in chapter 2.
Each move on the scales turns the equation into a new one, but they share a root: $3x + 2 = 14$, $3x = 12$ and $x = 4$ are true for the same $x$ and false for all others.
Equations are called equivalent if they have the same roots. Solving an equation is a chain of equivalent equations, the last of which is so simple that its root is obvious.
The scales suggest two rules, but scales are only a picture. Let's state the rules as statements about numbers and prove them from the laws of arithmetic of the last chapter.
If the same number is added to both sides of an equation or subtracted from both sides, the result is an equation equivalent to the original. The same holds if you add or subtract an expression in the unknown that makes sense for every value of the unknown, such as $2x$.
The idea: show that every root of one equation is a root of the other, and vice versa. Suppose the equation $A = B$ (where $A$ and $B$ are expressions in $x$) has been turned into $A + c = B + c$.
Take any root of the first equation, that is, a number $x_0$ for which $A$ and $B$ are the same number. Adding the same $c$ to the same number gives the same result. So $A + c = B + c$ is true at $x_0$ too: a root of the first equation is a root of the second.
Conversely, suppose $A + c = B + c$ is true at $x_0$. Add $-c$ to both sides. On the left we get $A + c + (-c) = A + \bigl(c + (-c)\bigr) = A + 0 = A$ by the associative law, and on the right, likewise, $B$. So $A = B$ at $x_0$: a root of the second equation is a root of the first. The roots coincide; the equations are equivalent. Subtracting $c$ is adding $-c$, and the argument is the same. If $c$ is an expression in $x$ defined for all $x$, then at each $x_0$ it becomes a number, and the argument goes through unchanged.
If both sides of an equation are multiplied or divided by the same number $k \ne 0$, the result is an equation equivalent to the original. For $k = 0$ this is false.
The argument is the same as for the first rule. If the sides $A$ and $B$ are equal at $x_0$, then so are $kA$ and $kB$, products of equal numbers. Conversely, suppose $kA = kB$ at $x_0$. Since $k \ne 0$, it has a reciprocal $\frac1k$; multiply both sides by it. On the left, by the associative law, $\frac1k \cdot (kA) = \bigl(\frac1k \cdot k\bigr)A = 1 \cdot A = A$, and on the right, likewise, $B$. So $A = B$ at $x_0$, and the roots coincide. Dividing by $k$ is multiplying by $\frac1k$, which isn't zero either.
The condition $k \ne 0$ can't be dropped: zero has no reciprocal, and the reverse step fails. The equation $x + 1 = 3$ has the single root $2$, but multiplying by zero gives $0 = 0$, for which any number is a root. Equivalence is lost.
Unlike the first rule, the second can't be applied carelessly to expressions in the unknown: such an expression may become zero for some $x$. What happens then we'll see in the section on traps.
Completion and balancing
Around 820 in Baghdad, Muhammad ibn Musa al-Khwarizmi, a scholar of the House of Wisdom at the court of the caliph al-Ma'mun, wrote The Compendious Book on Calculation by Completion and Balancing. It contains not a single letter or symbol. The unknown is called a "thing", its square "wealth", and numbers "dirhams", after the coin. The equation $x^2 + 10x = 39$ reads, in his words: wealth and ten roots are equal to thirty-nine dirhams. We'll solve such equations in chapter 10; for now we're interested in the two operations in the book's title.
Al-jabr, completion, rids an equation of something subtracted: you add what was subtracted to both sides. From $4x - 3 = 2x + 5$ you get $4x = 2x + 8$. On our scales, that means tying a weight to each pan for every balloon. Al-muqabala, balancing, removes the same thing from both sides: from $4x = 2x + 8$ you get $2x = 8$. On the scales, that means taking one box off each pan.
Why did al-Khwarizmi need completion? He didn't accept negative numbers, and an equation counted as put in order only when both sides contained nothing but things being added. So he had six separate kinds of equation, such as "wealth and roots equal a number" or "wealth and a number equal roots", each with its own recipe. We have it easier: with negative numbers all six kinds merge into one.
Today both operations are written as one rule.
A term can be moved from one side of an equation to the other by changing its sign: the result is an equivalent equation.
Suppose the left side contains a term $c$: the equation has the form $A + c = B$. Subtract $c$ from both sides; by the first rule of balance we get the equivalent equation $A + c - c = B - c$. On the left $c - c = 0$, leaving $A = B - c$: the term $c$ has vanished from the left and appeared on the right with a minus sign. If the term had a minus sign, $A - c = B$, add $c$ to both sides to get $A = B + c$. That is exactly al-muqabala and al-jabr written in one line.
In $4x - 3 = 2x + 5$ move $-3$ to the right, where it becomes $+3$, and $2x$ to the left, where it becomes $-2x$: $4x - 2x = 5 + 3$, that is, $2x = 8$ and $x = 4$. Al-Khwarizmi's two operations fit into one line.
The three fates of a linear equation
Whatever brackets and fractions an equation contains, if the unknown occurs only to the first power and not in a denominator, then after expanding brackets and moving terms it becomes $ax + b = 0$.
An equation of the form $ax + b = 0$, where $a$ and $b$ are numbers, is called linear. The name comes from geometry: in chapter 9 we'll see that the expression $ax + b$ draws a straight line.
If $a \ne 0$, there is exactly one root. But $a$ may turn out to be zero. On the scales that means both pans held the same number of boxes, and after al-muqabala none are left. The equation becomes $0 \cdot x = -b$, and then there are two possible outcomes. If $b \ne 0$, the left side is zero for every $x$ while the right isn't: there are no roots, and no box can balance the scales. If $b = 0$, we get $0 \cdot x = 0$, and every number is a root: the scales don't care what the box weighs.
The equation $ax + b = 0$ has the single root $x = -\frac{b}{a}$ when $a \ne 0$; it has no roots when $a = 0$ and $b \ne 0$; and every number is a root when $a = 0$ and $b = 0$.
The idea: move $b$ to the right and see whether you can divide by $a$. By the rule for moving terms, $ax + b = 0$ is equivalent to $ax = -b$.
Let $a \ne 0$. By the second rule of balance, divide both sides by $a$ to get the equivalent equation $x = -\frac{b}{a}$. It has exactly one root, the number $-\frac{b}{a}$ itself, and equivalent equations have the same roots. So the original has one root too.
Let $a = 0$. Then for every $x$ the left side $0 \cdot x$ is zero (by the lemma on multiplying by zero from chapter 2), and the equation says $0 = -b$. If $b \ne 0$, this is false for every $x$: no roots. If $b = 0$, it's true for every $x$: every number is a root. There are no other possibilities: $a$ is either zero or not, and so is $b$.
The weighing log shows why there are exactly three cases. The tilt of the scales changes by the same step each time: every extra kilogram in the box shifts it by the same amount. So the points lie on a straight line. A sloping line crosses the zero level exactly once. A horizontal line either runs above or below zero and never meets it, or coincides with it entirely. There is no fourth option: a linear equation never has exactly two roots.
How many roots does the equation $2(x - 1) = 2x - 2$ have?
Expanding the brackets gives $2x - 2 = 2x - 2$, and moving terms gives $0 \cdot x = 0$. It's an identity: every number is a root.
Brackets and fractions
So little is known about the life of Diophantus, the man who gave the unknown its own sign, that even his century is inferred from indirect clues. But the Greek Anthology, an old collection of short poems, contains an epitaph-puzzle about his age. Retold, it goes like this. Diophantus's childhood took up a sixth of his life, his youth another twelfth. After another seventh of his life he married, and five years later a son was born. The son lived half as long as his father, and four years after the son's death Diophantus died. How long did he live?
Who wrote the epitaph, and when, is unknown, and it shouldn't be trusted as a biography. But as a puzzle it's lovely. Suppose Diophantus lived $x$ years; all the stretches of his life together make up his whole life:
$$\frac{x}{6} + \frac{x}{12} + \frac{x}{7} + 5 + \frac{x}{2} + 4 = x.$$The fractions get in the way, but the second rule of balance gets rid of them. The least common multiple of the denominators 6, 12, 7 and 2 is 84; remember the lcm from chapter 4. Multiply both sides by 84, that is, every term:
$$14x + 7x + 12x + 420 + 42x + 336 = 84x.$$Move the terms with $x$ to the left and the numbers to the right, and collect like terms: $75x - 84x = -756$, that is, $-9x = -756$ and $x = 84$. Childhood lasted 14 years, youth ended at 21, Diophantus married at 33, his son was born when he was 38 and lived 42 years, and Diophantus himself died at 84.
Every linear equation is solved by the same sequence of steps:
- clear the fractions by multiplying both sides by a common denominator;
- expand the brackets;
- move the terms with the unknown to one side and the numbers to the other;
- collect like terms to get $ax = c$;
- divide by $a$ if $a \ne 0$, or recognise one of the two special cases;
- check the answer by substituting it into the original equation.
The last step isn't a formality: it catches mistakes made at any of the earlier ones. In the widget below you can type your own equation and watch this sequence at work on it.
Solve $\frac{2x - 1}{3} = \frac{x + 4}{2}$.
The common denominator is 6. Multiply both sides by 6: $2(2x - 1) = 3(x + 4)$, that is, $4x - 2 = 3x + 12$. Move terms: $4x - 3x = 12 + 2$, $x = 14$. Check: $\frac{28 - 1}{3} = 9$ and $\frac{14 + 4}{2} = 9$.
Solve $0.3(x - 2) = 0.2x + 0.4$.
Decimals are just fractions with denominator 10. Multiply both sides by 10: $3(x - 2) = 2x + 4$, $3x - 6 = 2x + 4$, $x = 10$. Check: $0.3 \cdot 8 = 2.4$ and $0.2 \cdot 10 + 0.4 = 2.4$.
For harder equations, where it helps to see each step with an explanation, there's a linear equation solver.
Where the scales deceive
Let's solve $x^2 = x$. Divide both sides by $x$ and we get $x = 1$. Check: $1^2 = 1$, true. Is that all? No: $x = 0$ works too, since $0^2 = 0$. The root was lost in the division. The second rule of balance allows dividing only by a number other than zero, and $x$ might be exactly zero, the very case we threw away along with the root.
The right way is to move everything to one side and factor, as in the last chapter: $x^2 - x = 0$, $x(x - 1) = 0$. From there a simple but very useful statement takes over.
A product of two numbers is zero if and only if at least one of them is zero: $pq = 0 \iff p = 0$ or $q = 0$.
One way: if $p = 0$ or $q = 0$, then $pq = 0$, because multiplying by zero gives zero (chapter 2). The other way: suppose $pq = 0$ and $p \ne 0$. Then $p$ has a reciprocal, and by the associative law $q = 1 \cdot q = \bigl(\frac1p \cdot p\bigr)q = \frac1p \cdot (pq) = \frac1p \cdot 0 = 0$. So if the first factor isn't zero, the second is.
For $x(x - 1) = 0$ the lemma gives $x = 0$ or $x - 1 = 0$, that is, $x = 0$ or $x = 1$. Both roots are in place.
Never divide both sides of an equation by an expression in the unknown: at the values where it is zero, roots get lost. Instead of dividing, move everything to one side and factor.
Multiplying by an expression in the unknown is dangerous in a different way: it can add a root that wasn't there. Take $\frac{x}{x - 2} = \frac{2}{x - 2}$ and multiply both sides by $x - 2$: we get $x = 2$. But at $x = 2$ the denominators become zero, and the original equation makes no sense. It has no roots; the two turned up because at $x = 2$ we were multiplying by zero.
An extraneous root is a number that turns up during solving but doesn't satisfy the original equation. Extraneous roots come from non-equivalent transformations: multiplying by an expression in the unknown, and later squaring both sides. That's why an answer is always checked by substituting it into the original equation.
Below are seven solutions. Each has exactly one wrong step, and dividing by an expression in the unknown isn't the only kind. Try to find the mistake before you read the explanation.
A problem written in words
Most equations in life come not as a formula but as a story. Translating from words into an equation takes four moves. Decide what the unknown will be and give it a letter. Express every quantity mentioned in the problem in terms of it. Find a quantity that can be computed in two ways, and set the two equal. Once the equation is solved, translate the answer back into words and ask yourself whether it's reasonable.
Two villages are 36 kilometres apart. At the same moment a cyclist sets off from one towards the other at 13 km/h, and a walker from the other towards the first at 5 km/h. When will they meet? Say after $x$ hours. By then the cyclist has ridden $13x$ kilometres and the walker has walked $5x$, and together they've covered the whole distance: $13x + 5x = 36$. So $18x = 36$ and $x = 2$. After two hours the cyclist is 26 kilometres from his village and the walker 10 kilometres from hers, and $26 + 10 = 36$.
The answer to an equation deserves a careful reading. A father is 40, his son 10. In how many years will the father be twice as old as the son? $40 + x = 2(10 + x)$, so $x = 20$: the father will be 60 and the son 30. And five times as old? $40 + x = 5(10 + x)$ gives $x = -2.5$. A negative answer isn't a mistake: it means two and a half years ago, when the father was $37.5$ and the son $7.5$. The equation doesn't know the words "in" and "ago"; it simply shows you where to look.
A bat and a ball cost a dollar ten in total. The bat costs a dollar more than the ball. How much does the ball cost?
Let the ball cost $x$ cents; then the bat costs $x + 100$, and $x + (x + 100) = 110$. So $2x = 10$ and $x = 5$. Intuition suggests "10 cents", and the check refutes it.
The sum of three consecutive natural numbers is 96. Find the smallest of them.
Let the smallest number be $x$; then the other two are $x + 1$ and $x + 2$. We get $x + (x + 1) + (x + 2) = 96$, $3x + 3 = 96$, $x = 31$. Check: $31 + 32 + 33 = 96$.
Where next
The scales answered the question of how much the box weighs, and the answer was a number. But go back to the cyclist and the walker. The distance between them isn't one number: an hour after the start it's 18 kilometres, after half an hour 27, and after $t$ hours $36 - 18t$. The equation asked when this distance becomes zero. What if you need to see the whole story at once: how one quantity depends on another, where it grows, where it shrinks, when it hits zero? Then the answer isn't a number but a rule. Such a rule is called a function, and the points in the weighing log have already hinted at how to draw it. That's chapter 9.