Linear equations and inequalities

Lens

MotivationH

Two tanks

At a school in Lilongwe, two rainwater tanks stand side by side. In the morning the first holds $500$ litres and is being used to water the vegetable garden, losing $20$ litres a minute. The second holds $100$ litres and is being filled from the roof gutters at $30$ litres a minute. When will the two tanks hold the same amount?

After $t$ minutes the first tank holds $500 - 20t$ litres and the second $100 + 30t$ litres. We want the value of $t$ that makes

500 - 20t = 100 + 30t.

We could guess: after $5$ minutes the tanks hold $400$ and $250$ litres; after $10$ minutes, $300$ and $400$ . So the answer lies somewhere between. Guessing works here, but it is slow, it is hopeless when the answer is $\tfrac{37}{11}$ , and it can never tell us that there is no answer. This lesson develops a method that always works — and proves why it works.

IntuitionH

Keeping the balance

An equation is like a balance in equilibrium: the two sides weigh the same. If we take the same amount off both pans, or double both pans, the balance stays level. Try it: each box weighs $x$ units, and the goal is to get one box alone on one side.

x on both sides

Get one x alone. Every legal move keeps the scale level.

x (unknown weight)
1 (unit weight)

3x + 1 = x + 9

Do the same thing to both sides. Try to get x on its own.

The scale is still level: both sides changed by the same amount.

Your steps

3x + 1 = x + 9

Make your own equation

Left side

Number of x's3

Number of 1s1

Right side

Number of x's1

Number of 1s9

DefinitionsH

Equations, solutions and equivalence

Definition— equation and solution

An equation in a variable $x$ is a statement that two algebraic expressions are equal, such as $3x + 1 = x + 9$ . A solution is a value of $x$ that makes the statement true. The solution set is the set of all solutions. To solve an equation means to find its solution set — all of the solutions, and nothing else.

Definition— equivalent equations

Two equations are equivalent if they have exactly the same solutions.

Definition— linear equation

A linear equation in $x$ is one that can be written as $ax + b = cx + d$ for numbers $a, b, c, d$ : the variable appears only to the first power, multiplied by numbers.

So $5 - 2x = 11$ and $\tfrac{x}{3} + \tfrac{1}{2} = \tfrac{x - 1}{4}$ are linear; $x^2 = 2x$ and $\tfrac{1}{x} = 3$ are not.

Definition— inequality

An inequality compares two expressions with $<$ , $\leq$ , $>$ or $\geq$ . Its solutions are the values of the variable that make it true. For example, the solutions of $x + 2 > 5$ are all the numbers greater than $3$ — infinitely many.

Throughout, the variable can be any rational number (see rational numbers) unless we say otherwise. This matters: $2x = 1$ has the solution $\tfrac{1}{2}$ among the rational numbers but no solution among the integers.

ProofH

Why the balance method works

Theorem— moves that give equivalent equations

For all numbers $p$ , $q$ and $c$ :

$p = q$ if and only if $p + c = q + c$ ;
if $c \neq 0$ , then $p = q$ if and only if $cp = cq$ .

Consequently, adding the same number to both sides of an equation, or multiplying both sides by the same non-zero number, gives an equivalent equation.

Proof

If $p = q$ , then $p + c$ and $q + c$ are the same number, so they are equal. Conversely, if $p + c = q + c$ , add $-c$ to both sides: $p = (p + c) + (-c) = (q + c) + (-c) = q$ .
If $p = q$ , then $cp = cq$ . Conversely, if $cp = cq$ and $c \neq 0$ , multiply both sides by $\tfrac{1}{c}$ , which exists because $c \neq 0$ : $p = \tfrac{1}{c}(cp) = \tfrac{1}{c}(cq) = q$ .

For an equation in $x$ , apply this to each value of $x$ separately, with $p$ and $q$ the values of the two sides: $x$ makes the old equation true exactly when it makes the new one true.

∎

Subtracting $c$ is adding $-c$ , and dividing by $c \neq 0$ is multiplying by $\tfrac{1}{c}$ , so these moves are covered too. We may also add an expression such as $-x$ to both sides: for each value of $x$ , that just adds a number.

The proof shows exactly where $c \neq 0$ is used: to undo the multiplication. Multiplying by $0$ cannot be undone. The equation $x = 3$ has one solution, but $0 \cdot x = 0 \cdot 3$ is true for every $x$ — multiplying by zero has thrown the information away.

So solving an equation is building a chain of equivalent equations, ending with one whose solutions we can read off. Because every link is an "if and only if", the last equation has exactly the same solutions as the first. Checking the answer by substitution is still a good habit: it catches arithmetic slips.

Theorem— the three outcomes for ax = b

For numbers $a$ and $b$ , the equation $ax = b$ has

exactly one solution, $x = \tfrac{b}{a}$ , if $a \neq 0$ ;
no solution, if $a = 0$ and $b \neq 0$ ;
every number as a solution, if $a = 0$ and $b = 0$ .

Proof

If $a \neq 0$ , multiplying both sides by $\tfrac{1}{a}$ gives the equivalent equation $x = \tfrac{b}{a}$ , whose only solution is $\tfrac{b}{a}$ . If $a = 0$ , the left side $0 \cdot x$ is $0$ for every $x$ , so the equation says $0 = b$ : never true if $b \neq 0$ , always true if $b = 0$ .

∎

Every linear equation can be brought into this form. Subtracting $cx + b$ from both sides of $ax + b = cx + d$ gives the equivalent equation

(a - c)\,x = d - b.

Theorem— no linear equation has exactly two solutions

If two different numbers $x_1 \neq x_2$ both solve $ax + b = cx + d$ , then every number solves it.

Proof

Both numbers solve the standard form: $(a - c)x_1 = d - b$ and $(a - c)x_2 = d - b$ . Subtracting, $(a - c)(x_1 - x_2) = 0$ . Since $x_1 - x_2 \neq 0$ , we can divide by it, so $a - c = 0$ , and then $d - b = (a - c)x_1 = 0$ . So the standard form reads $0 \cdot x = 0$ , which every number satisfies.

∎

A linear equation therefore has exactly one solution, no solution, or infinitely many — nothing in between.

Worked examplesH

Worked examples

Example— The variable on both sides

Solve $3x + 1 = x + 9$ .

\begin{aligned} 3x + 1 &= x + 9 &&\\ 2x + 1 &= 9 &&\text{subtract } x \text{ from both sides}\\ 2x &= 8 &&\text{subtract } 1\\ x &= 4 &&\text{multiply by } \tfrac{1}{2} \end{aligned}

Check: $3 \cdot 4 + 1 = 13$ and $4 + 9 = 13$ . ✓

Example— The two tanks

$500 - 20t = 100 + 30t$ . Add $20t$ to both sides: $500 = 100 + 50t$ . Subtract $100$ : $400 = 50t$ . Divide by $50$ : $t = 8$ . After $8$ minutes both tanks hold $340$ litres ( $500 - 160 = 100 + 240$ ).

Example— Negative coefficients and brackets

$5 - 2x = 11$ . Subtract $5$ : $-2x = 6$ . Divide by $-2$ : $x = -3$ . Check: $5 - 2(-3) = 5 + 6 = 11$ . ✓
$2(x - 3) - (x - 5) = 7$ . Expand carefully — the minus sign in front of the second bracket multiplies both terms: $2x - 6 - x + 5 = 7$ , so $x - 1 = 7$ and $x = 8$ . Check: $2 \cdot 5 - 3 = 7$ . ✓

Example— Fractions

Solve $\dfrac{x}{3} + \dfrac{1}{2} = \dfrac{x - 1}{4}$ .

Multiply both sides by $12$ , the least common multiple of $3$ , $2$ and $4$ — and multiply every term:

12 \cdot \frac{x}{3} + 12 \cdot \frac{1}{2} = 12 \cdot \frac{x - 1}{4} \quad\Longrightarrow\quad 4x + 6 = 3x - 3 \quad\Longrightarrow\quad x = -9.

Check: $\tfrac{-9}{3} + \tfrac{1}{2} = -\tfrac{5}{2}$ and $\tfrac{-10}{4} = -\tfrac{5}{2}$ . ✓

Example— No solution, and every number a solution

$2(x + 3) = 2x + 5$ expands to $2x + 6 = 2x + 5$ . Subtracting $2x$ leaves $6 = 5$ , which is false whatever $x$ is. No solution.
$3(x - 2) + x = 4x - 6$ expands to $4x - 6 = 4x - 6$ . Subtracting $4x$ leaves $-6 = -6$ , true whatever $x$ is. Every number is a solution: the equation is an identity.

When the variable disappears, do not panic — read the statement that remains.

Example— Modelling: three consecutive numbers

Three consecutive integers add up to $96$ . What are they? Call the smallest $n$ ; the others are $n + 1$ and $n + 2$ . Then $n + (n + 1) + (n + 2) = 96$ , so $3n + 3 = 96$ , $3n = 93$ and $n = 31$ . The numbers are $31, 32, 33$ — and indeed they add to $96$ . Choosing what the variable stands for is the most important step in modelling.

ProofH

Inequalities, and why they flip

Inequalities are solved with the same moves — with one twist. We use the basic facts about order: $p < q$ means that $q - p$ is positive; the sum and the product of two positive numbers are positive; and the product of a positive and a negative number is negative.

Theorem— moves for inequalities

For all numbers $p$ , $q$ and $c$ :

$p < q$ if and only if $p + c < q + c$ ;
if $c > 0$ , then $p < q$ if and only if $cp < cq$ ;
if $c < 0$ , then $p < q$ if and only if $cp > cq$ — the inequality reverses.

The same holds with $\leq$ in place of $<$ .

Proof

$(q + c) - (p + c) = q - p$ , so one difference is positive exactly when the other is.
If $c > 0$ and $q - p > 0$ , then $cq - cp = c(q - p)$ is a product of positive numbers, so it is positive: $cp < cq$ .
If $c < 0$ and $q - p > 0$ , then $-c > 0$ , so $(-c)(q - p) > 0$ . But $(-c)(q - p) = cp - cq$ . So $cp - cq > 0$ , which says $cq < cp$ .

Each move can be undone (by adding $-c$ , or multiplying by $\tfrac{1}{c}$ , which has the same sign as $c$ ), which gives the "if and only if".

∎

Here is the picture behind part 3. Multiplying by $-1$ reflects the number line in $0$ : $2 < 5$ , but $-2 > -5$ , because the reflection swaps left and right. Multiplying by any negative number is a stretch followed by this reflection.

Example— Solving an inequality

Solve $7 - 3x \leq 19$ .

\begin{aligned} 7 - 3x &\leq 19 &&\\ -3x &\leq 12 &&\text{subtract } 7\\ x &\geq -4 &&\text{divide by } -3 \text{, which reverses the inequality} \end{aligned}

Check with test values: $x = 0$ gives $7 \leq 19$ , true; $x = -4$ gives $19 \leq 19$ , true; $x = -5$ gives $22 \leq 19$ , false. ✓

The solution set of 7 − 3x ≤ 19 is x ≥ −4: the filled dot shows that −4 itself is included, and the arrow shows that every larger number is a solution too.

Example— Modelling: how many notebooks?

Mariam, a student in Fez, has $200$ dirhams. She needs a book costing $45$ dirhams and wants as many notebooks as possible at $12$ dirhams each. With $n$ notebooks she needs $45 + 12n \leq 200$ , so $12n \leq 155$ and $n \leq 12.91\ldots$ But $n$ counts notebooks, so it is a whole number: the solutions are $n = 0, 1, \ldots, 12$ , and she can buy at most $12$ . (Thirteen would cost $45 + 156 = 201$ dirhams — one too many!)

Formal mathematicsF

Equivalent equations in Lean

In Lean we work with the rational numbers Rat. "Two equations are equivalent" becomes a statement with ↔ ("if and only if") that holds for every x. First the two basic moves; each proof undoes the move, just as in the human proof:

add_both_sidesLean 4Proved here from basic factsFormally verified

/-- Adding the same number `c` to both sides gives an equivalent equation:
`p = q ↔ p + c = q + c`. Forwards, equal numbers plus `c` are equal. Backwards, the move can
be **undone**: subtract `c` again, `p = (p + c) − c = (q + c) − c = q`. -/
theorem add_both_sides (p q c : Rat) : p = q ↔ p + c = q + c := by
  constructor
  · intro h
    rw [h]
  · intro h
    calc p = p + c - c := Rat.add_sub_cancel.symm
      _ = q + c - c := by rw [h]
      _ = q := Rat.add_sub_cancel