A taxi charges $3.00 to start the meter plus $2.50 per mile. Your ride cost $23.00. How far did you travel?

Call the distance \(d\) miles. Then the cost is \(3 + 2.5d\), and the information you have is

You can probably see the answer quickly: take off the $3 start fee to leave $20, and $20 at $2.50 a mile is 8 miles. That two-step mental calculation is exactly the algorithm this chapter formalises. Before we do that, let us name what kind of equation this is, because the name tells you which tools apply.

Which of these are linear in \(x\)?

"I think of a number, multiply it by 3, subtract 5, and get 16. What was my number?"

Written as an equation, \(3x - 5 = 16\). The puzzle describes a little pipeline of operations applied to the unknown:

To recover \(x\) you run the pipeline backwards, undoing each stage with its inverse, last stage first:

The number was 7. Check: \(3 \cdot 7 - 5 = 16\). That is the entire idea of solving a linear equation. Everything else is bookkeeping for when the pipeline is messier: when \(x\) appears on both sides, inside brackets, or inside fractions.

You put on socks, then shoes. To undo, you take off shoes first, then socks. Inverse operations undo in reverse order for the same reason: the last thing done to \(x\) is the outermost layer, so it has to come off first. In \(3x - 5\), the subtraction of 5 happened after the multiplication by 3 (order of operations), so it is undone first.

The second intuition is the balance from Chapter 4: an equation is a pair of scales in equilibrium. Whatever you do to one pan, you must do to the other, or the balance tips and the statement is no longer the one you started with.

Why is it legitimate to "do the same thing to both sides"? Suppose \(x = s\) is a solution of \(L = R\), meaning that when \(x\) is replaced by \(s\) the two sides are the same number. Adding 5 to two equal numbers gives two equal numbers; multiplying two equal numbers by 3 gives two equal numbers. So every solution of the old equation is a solution of the new one.

That is only half of what we need. We also need the reverse: the new equation should not have gained solutions. That is where the inverse matters. If you added 5, you can subtract 5 to get back exactly the original equation, so any solution of the new equation is also a solution of the old one. Each step is reversible, and reversible steps produce equivalent equations.

Now you can see why multiplying both sides by zero is forbidden. Multiplying \(2x = 6\) by 0 gives \(0 = 0\), which is true for every \(x\). Information has been destroyed and cannot be recovered, because there is no "divide by 0" to undo it. The two legal moves, stated precisely:

The double arrow \(\Longleftrightarrow\) means "if and only if": each equation implies the other, so their solution sets are identical. Subtraction is adding \(-k\); division is multiplying by \(1/k\). Simplifying a side (expanding brackets, collecting like terms, as in Chapter 4) is the third legal move, since it does not change the value of that side at all.

Example 1. Solve \(3(2x - 5) = 4x + 7\).

Check: left side \(3(2 \cdot 11 - 5) = 3 \cdot 17 = 51\); right side \(4 \cdot 11 + 7 = 51\). Equal, so \(x = 11\).

Example 2. Solve \(5 - 2(x - 3) = 3x - 4\).

The danger here is the minus sign in front of the bracket. \(-2(x - 3)\) means \(-2\) multiplied by each term inside: \((-2)(x) + (-2)(-3) = -2x + 6\).

Check: left \(5 - 2(3 - 3) = 5 - 0 = 5\); right \(3 \cdot 3 - 4 = 5\). Equal.

Notice the choice made at the fourth line. Adding \(2x\) to both sides moved the \(x\)-terms to the right, where they ended up with the positive coefficient \(5\). Subtracting \(3x\) instead would have given \(11 - 5x = -4\), then \(-5x = -15\), then \(x = 3\): the same answer, but with two extra negative signs to get wrong. An equation with \(x\) on the right, like \(3 = x\), is perfectly fine; equality is symmetric.

For one-off equations by hand, use the procedure: it never fails and every step is checkable. Use the reverse flowchart when \(x\) appears once; it is quicker and harder to get wrong. Use a graph when you want to see the solution or to check a suspicious answer. Use guess-and-check only when the equation is not linear and no algebraic method is available.

Any time a formula gives you the answer to the wrong question, you solve a linear equation to turn it round.

Problems about shares and proportions ("a third of one amount, less a quarter of another") arrive full of fractions. A typical result is

You could work with the fractions throughout, but there is a better move. Every term can be multiplied by the same nonzero number without changing the solution set, so multiply by a number that cancels every denominator at once.

Solve \(\dfrac{x + 1}{3} - \dfrac{x - 2}{4} = 1\). The LCD of 3 and 4 is 12.

Check: \(\frac{2 + 1}{3} - \frac{2 - 2}{4} = 1 - 0 = 1\). Correct.

The single most common error here is in line 2. The fraction bar is a grouping symbol: \(\frac{x - 2}{4}\) means \(\frac{1}{4}(x - 2)\), so after clearing, the \(-3\) multiplies the whole of \((x - 2)\). Writing the brackets explicitly at this stage, before expanding, prevents the mistake.

Decimals are fractions in disguise, cleared by a power of 10. For \(0.05x + 0.12(200 - x) = 17\), multiply through by 100:

Check: \(0.05(100) + 0.12(100) = 5 + 12 = 17\).

Strictly, \(\frac{2}{x} + 3 = \frac{5}{x}\) is not linear, but it becomes linear after multiplying by \(x\). The catch is that \(x\) could be zero, and multiplying by zero is not reversible. So first record the restriction \(x \neq 0\), then proceed:

\(x = 1\) satisfies the restriction, and the check \(2 + 3 = 5\) confirms it.

Try to solve \(3x + 1 = 3x + 4\). Subtract \(3x\) from both sides and you get \(1 = 4\). The \(x\) has vanished and left a false statement. What went wrong?

Nothing went wrong. The equation asks "which numbers \(x\) make \(3x + 1\) equal to \(3x + 4\)?" and the answer is "none": whatever \(x\) is, the right side is always 3 bigger. The algebra correctly reported that.

After expanding and collecting, every linear-looking equation \(ax + b = cx + d\) becomes

Call the coefficient \(A = a - c\) and the right side \(B = d - b\). Then:

Example. Solve \(6x - 2(3x + 1) = -2\).

This is always true, so the solution set is \(\mathbb{R}\): every real number works. Test any value you like; \(x = 100\) gives \(600 - 602 = -2\).

The picture behind these cases is simple and will matter a great deal in Chapter 6. Graph each side as a line, \(y = ax + b\) and \(y = cx + d\). The \(x\)-coefficients \(a\) and \(c\) are the slopes. Different slopes: the lines cross exactly once. Equal slopes, different intercepts: parallel lines that never meet. Equal slopes and equal intercepts: the same line, meeting everywhere.

Back in the taxi: you have $25. How far can you go? Now the question is not "which distance costs exactly $25" but "which distances cost at most $25":

The answer is not a single number but a whole range of numbers. Inequalities are how mathematics talks about budgets, limits, tolerances, capacities and thresholds: anything with "at most", "at least", "no more than", "exceeds".

Solving the taxi inequality looks exactly like solving the equation:

Any distance up to 8.8 miles (and, since distance cannot be negative, at least 0) is affordable. So far, inequalities behave like equations. There is exactly one twist.

Start with a true statement, \(2 < 4\), and multiply both sides by \(-1\). You get \(-2\) and \(-4\). Which is bigger? On the number line \(-2\) is to the right of \(-4\), so \(-2 > -4\). The inequality has reversed.

Geometrically, multiplying by \(-1\) is a mirror reflection of the number line about 0. A reflection turns left-to-right into right-to-left, so "is to the left of" becomes "is to the right of". Multiplying by any negative number is a reflection followed by a stretch, and the stretch does not change order, so every negative multiplier reverses the inequality.

Here is the argument in symbols; it is short and it shows exactly where the sign of the multiplier enters. Recall that \(a < b\) means the same as \(b - a > 0\) (the gap from \(a\) up to \(b\) is positive).

Suppose \(a < b\), so \(b - a > 0\), and let \(k\) be any number. Then

Adding the same number to both sides never changes the order, because \((b + k) - (a + k) = b - a\): the gap is unchanged. So the rules for inequalities are the rules for equations, plus one exception.

The same rules hold with \(\le\) in place of \(<\). Swapping the two sides also reverses the symbol: \(3 < x\) says the same thing as \(x > 3\).

Solve \(7 - 3x \le 19\).

Check with a value inside, \(x = 0\): \(7 - 0 = 7 \le 19\) is true. A value outside, \(x = -5\): \(7 + 15 = 22 \le 19\) is false. The boundary, \(x = -4\): \(7 + 12 = 19 \le 19\) is true, so the endpoint belongs to the solution set, as \(\ge\) says. Solution set: \(\lbrack -4, \infty)\).

You can sidestep the sign reversal by moving the \(x\)-term to whichever side makes its coefficient positive:

\(-4 \le x\) is the same statement as \(x \ge -4\). Many people find this method more reliable, because it replaces a rule you have to remember with a move you cannot get wrong. Use whichever you prefer; but know both, because when the coefficient is a symbol \(a\) whose sign you do not know, you need to split into cases \(a > 0\) and \(a < 0\) and the flip rule is unavoidable.

A solution like "all \(x\) with \(-2 \le x < 3\)" is common enough to deserve a compact notation.

A compound inequality joins two inequalities with "and" or "or".

"And": a double inequality. \(-3 < 2x + 1 \le 9\) means \(-3 < 2x + 1\) and \(2x + 1 \le 9\). Because the middle is sandwiched, you can apply each operation to all three parts at once:

Solution set \((-2, 4]\). Check the endpoints: \(x = -2\) gives \(-3 < -3\), false, so excluded; \(x = 4\) gives \(9 \le 9\), true, so included.

When the multiplier is negative, both symbols flip, and it is tidiest to rewrite the result so the smaller number is on the left:

"Or": a union. Solve \(3x - 1 < -7\) or \(2x + 5 \ge 11\). Solve each separately: \(3x < -6\) gives \(x < -2\); \(2x \ge 6\) gives \(x \ge 3\). A number satisfies the compound statement if it satisfies either part, so the solution set is the union

"And" corresponds to intersection (the overlap of the two solution sets); "or" corresponds to union (everything covered by either). An "and" can be empty: \(x < 3\) and \(x > 5\) share no numbers, so its solution set is \(\varnothing\).

Inequalities answer "how much can I afford?", "how many will fit?" and "what do I need?".

A machined bolt should be 25.00 mm long, with a tolerance of 0.05 mm. Which lengths \(L\) pass inspection? In words: "\(L\) is within 0.05 of 25". In symbols:

Absolute values are the language of distance, error and tolerance. They look awkward because the absolute-value function is not linear (it has a corner), but every absolute-value equation or inequality with a linear inside splits into linear pieces you already know how to solve.

Read every absolute-value statement as a distance statement and the answer is almost visible:

"Less than" gives one interval in the middle (an "and"); "greater than" gives two outer rays (an "or"). This is the whole theory, and it follows directly from the definition: \(|u| = k\) with \(k > 0\) means \(u\) is either \(k\) or \(-k\); \(|u| < k\) means \(u\) is between \(-k\) and \(k\).

The same holds with \(\le\) and \(\ge\). If \(k < 0\), \(|u| = k\) and \(|u| < k\) have no solutions (a distance cannot be negative), while \(|u| > k\) is true for every \(u\).

Example 1. Solve \(3|x - 1| + 2 = 14\).

Check: \(3|4| + 2 = 14\) and \(3|-4| + 2 = 14\). Both work. (Geometrically: the two points at distance 4 from 1.)

Example 2. Solve \(|2x - 5| = 7\).

\(2x - 5 = 7\) gives \(x = 6\); \(2x - 5 = -7\) gives \(2x = -2\), \(x = -1\). Check: \(|7| = 7\) and \(|-7| = 7\).

Example 3. Solve \(|2x + 1| \ge 5\).

Solution set \((-\infty, -3] \cup [2, \infty)\). Test \(x = 0\), which should fail: \(|1| = 1 \ge 5\) is false. Good.

Example 4. The bolt: \(|L - 25| \le 0.05\) becomes \(-0.05 \le L - 25 \le 0.05\), so \(24.95 \le L \le 25.05\).

Choosing: use the distance reading for \(|x - a|\) forms, the split procedure for a single \(|ax + b|\) isolated on one side, and case analysis whenever \(x\) also appears outside the absolute value. Whenever \(x\) appears outside, check every candidate in the original equation, because the right side might be negative.

Any range \(\lbrack a, b]\) can be written as one absolute-value inequality: centre \(c = \frac{a + b}{2}\), half-width \(r = \frac{b - a}{2}\), and \(a \le x \le b \iff |x - c| \le r\). That is how specifications, tolerances and error bars are quoted.

So far the unknown has been a single number to be found. Step back and ask a bigger question about the taxi: how does the fare depend on the distance, for every distance? Tabulate it:

Each extra mile adds the same $2.50. That constant step is the defining feature of a linear relationship, and it is why the graph is a straight line. Every equation you solved earlier in this chapter was a question about a line; seeing the line makes the algebra visible.

A relationship \(y = mx + b\) has two numbers that say everything about it.

Rate of change is the idea to hold on to. Speed is a rate of change (miles per hour), price is a rate of change (dollars per kilogram), throughput is a rate of change (requests per second). A linear model says the rate is the same everywhere. Calculus (Chapter 17) is, in large part, about what to do when it is not.

Why does \(y = mx + b\) give a straight line, and why is the slope the same whichever two points you pick? Take any two points on the graph, at \(x_1\) and \(x_2\). Then

The \(b\) cancels and the ratio is always \(m\). So every "rise over run" triangle drawn on the graph has the same shape (the triangles are similar, in the language of Chapter 9), and a curve whose rise-over-run triangles all have the same shape cannot bend: it is a straight line. Conversely, any non-vertical line has a constant ratio \(\Delta y / \Delta x\), which is exactly what \(y = mx + b\) encodes.

The same line can be written in several forms, each convenient for a different job.

Point–slope form is just the slope definition rearranged: if \((x, y)\) is any other point on the line, then \(\frac{y - y_1}{x - x_1} = m\); multiply both sides by \((x - x_1)\).

Find the line through \((2, 1)\) and \((6, 7)\), and its intercepts.

Check with the other point: \(\tfrac32 \cdot 6 - 2 = 9 - 2 = 7\). Correct.

The \(y\)-intercept is \((0, -2)\). The \(x\)-intercept solves a linear equation:

In standard form, multiply \(y = \tfrac32 x - 2\) by 2 and rearrange: \(3x - 2y = 4\).

Here is the payoff: every technique in this chapter has a picture.

A slope is a rate: "so much output per unit of input". Once you look for it, it is everywhere.

Real problems do not arrive as equations. They arrive as sentences, tables, invoices, and requirements. The mathematical skill is translation: deciding what the unknown is, expressing every quantity in terms of it, and recognising which statement is the equation. The solving, by now, is the easy part.

Cyclist A leaves at 9:00 riding at 12 km/h. Cyclist B leaves from the same place at 9:30 on the same route at 18 km/h. When and where does B catch A?

Let \(t\) be the number of hours after 9:00. A has ridden for \(t\) hours and B for \(t - 0.5\) hours (30 minutes later, in hours). Distance is speed times time, and "catches" means equal distances:

Check: by then A has ridden 1.5 h at 12 km/h, i.e. 18 km; B has ridden 1 h at 18 km/h, i.e. 18 km. Interpret: B catches A at 10:30, 18 km from the start.

How many litres of a 30% acid solution must be added to 10 litres of a 10% solution to obtain a 25% solution?

Let \(x\) be the litres of 30% solution added. The key is to track the quantity that is conserved: pure acid.

Acid in = acid out:

Check: 40 litres containing \(1 + 9 = 10\) litres of acid is \(10/40 = 25\%\). Interpret: add 30 litres. The same table structure solves blends of coffee by price, alloys by purity, and investments by interest rate: a weighted average is a linear equation.

A mobile provider offers Plan A at $20 per month plus $3 per GB, and Plan B at $50 per month plus $1 per GB. When is each plan cheaper?

Let \(g\) be the GB used in a month. The costs are \(C_A = 20 + 3g\) and \(C_B = 50 + g\). The break-even point is where they are equal:

At 15 GB both cost \(20 + 45 = 65\) dollars. Plan A is cheaper when \(20 + 3g < 50 + g\), i.e. \(2g < 30\), i.e. \(g < 15\). Interpret: below 15 GB a month choose A, above 15 GB choose B.

This shape, a fixed cost plus a marginal cost per unit, is the single most common model in business, and the comparison is always the same: the plan with higher fixed cost and lower marginal cost wins at high volume. (Strictly, finding the crossing of two lines is a system of two equations; Chapter 6 develops that view.)

A weather station logs 12 °C at 10:00 and 20 °C at 14:00. What was the temperature at 11:30? An animation moves a sprite from \(x = 100\) at frame 0 to \(x = 400\) at frame 60. Where is it at frame 45? A lookup table gives a sensor’s calibration at a few points; what does a reading between two entries mean?

All three ask for a value between two known values, assuming the quantity changes at a constant rate in between. That is linear interpolation, known in graphics and game programming as lerp.

Interpolating means walking along the straight line from one known value \(a\) to the other, \(b\). Describe your position by the fraction of the way you have gone, \(t\): \(t = 0\) at the start, \(t = 1\) at the end, \(t = 0.5\) halfway. The total change is \(b - a\), so after a fraction \(t\) of the journey you have covered \(t(b - a)\) of it:

The two forms are equal by expanding: \(a + tb - ta = (1 - t)a + tb\). The second form shows lerp as a weighted average: weight \(1 - t\) on \(a\), weight \(t\) on \(b\), weights summing to 1. It works unchanged for points, colours, and vectors: lerp each coordinate.

The inverse question, "what fraction of the way is the value \(v\)?", is a linear equation in \(t\):

Combine the two and you can remap a value from one range \(\lbrack a, b]\) to another \(\lbrack c, d]\): find \(t\) in the first range, then lerp in the second.

Temperature at 11:30, given 12 °C at 10:00 and 20 °C at 14:00. Measure time in hours: \(x_0 = 10\), \(x_1 = 14\), \(x = 11.5\).

The estimate is 15 °C. This is exactly the same as finding the line through \((10, 12)\) and \((14, 20)\) (slope 2 °C per hour) and evaluating it at 11.5: \(12 + 2 \cdot 1.5 = 15\). Interpolation is just "the equation of the line through two points", packaged for repeated use.

remap is the Celsius–Fahrenheit conversion in disguise (0 °C ↦ 32 °F, 100 °C ↦ 212 °F) and the standard way to turn a raw sensor reading into physical units.

Linear interpolation is the simplest member of a family. When data are smooth and you have more than two points, polynomial or spline interpolation (Chapter 7, Chapter 32) follows curves better; when data are noisy, fitting a line by regression (Chapter 31) beats passing exactly through two noisy points. Choose lerp when the points are close together relative to how quickly the quantity curves, when speed matters (it is one multiply and two adds), or when "straight-line between keyframes" is the behaviour you actually want.