Chapter 6.5 — Exercise 6.4 — Applications

Nine Situations, Nine Equations

Every equation in this list reduces to one of two shapes: y = mx when the relationship is pure proportion with nothing fixed (double the input, double the output), or y = mx + c when some part of the cost or quantity is fixed regardless of the rest — a flat library charge, an auto's first-kilometre rate, water's freezing point sitting at 32°F rather than 0°F. Spotting which shape a situation fits, before writing a single equation, is often the fastest route to setting it up correctly: a pure-ratio wording like "the ratio of milk to water is 5:2" signals y = mx immediately, while any wording naming a starting amount, a fixed fee, or an initial value separate from the rate signals clearly that a constant term is on its way into the equation somewhere.

Setting Up the Auto Fare Equation

The auto-fare problem is genuinely worth deriving explicitly, step by step, since the wording "₹15 for the first km, ₹8 for each one after that" doesn't translate directly into a single clean term the way most of the other rows above did. For a total journey of x kilometres, the fare works out to 15 for that first kilometre alone, plus 8 for each one of the remaining (x − 1) kilometres travelled: y = 15 + 8(x − 1) = 15 + 8x − 8 = 8x + 7. Reading the finished graph at x = 4 gives y = 39, which matches a direct algebraic check performed separately: 15 + 8(3) = 15 + 24 = 39 exactly.

From the graph, a ₹55 fare corresponds to 6 km travelled, and travelling a full 7 km would cost ₹63 — both figures readable directly once the line has actually been drawn, with no equation-solving needed at the point of reading. It's worth noticing what the "+7" specifically represents once the equation is in its simplified y = 8x + 7 form: not a real, separately-charged fee anywhere in the problem, but simply what falls out algebraically once 15 + 8(x−1) is expanded and the −8 and +15 combine — a reminder that a constant term in a derived equation doesn't always correspond to an obvious, named quantity in the original situation.

A Ratio Problem: Hydrogen and Oxygen

Water's total molecular weight splits between its two constituent elements, hydrogen and oxygen, in a fixed ratio of 1:8. Calling the hydrogen quantity x and oxygen quantity y, the ratio x:y = 1:8 rearranges directly to y = 8x. If oxygen is 12 grams, the graph shows hydrogen at 1.5 grams; if hydrogen is 3/2 grams, oxygen reads as 12 grams — the same equation, just entered from either side depending on which quantity happens to already be known and which one is being asked for. This is a genuinely different way of using a graph than the auto-fare example: there, x meant kilometres and y meant rupees, always in that fixed order; here, the graph is read forward for one question and backward for the next, precisely because both x and y represent physical weights of the same kind, so nothing about the situation privileges reading in only one direction.

The Crossing Point of Two Temperature Scales

The Celsius-to-Fahrenheit equation, F = 9C/5 + 32, answers a question with a genuinely surprising twist: is there a temperature that reads the same number on both scales? Setting the two readings equal to one another (x = 9x/5 + 32, using the single letter x for both scales at once) and solving the resulting equation gives x = −40 as the only value where this can happen. At exactly −40°, the thermometer shows the identical number whether read in Celsius or Fahrenheit — the single one point where the two scales' otherwise entirely different lines happen to cross each other exactly, out of the infinitely many temperatures that could in principle have been checked.

30°C on this same graph reads as 86°F, and 95°F reads back as 35°C — both confirming the graph is being read correctly, since converting 35°C forward with the formula gives 9(35)/5 + 32 = 63 + 32 = 95, exactly matching the value originally started from. That two-way check — going forward with the formula, then backward again to confirm the graph agrees — is worth running whenever a graph-reading answer feels uncertain, since the formula and the picture are describing the exact same relationship and should never actually disagree once both are done correctly.

Why a Graph Beats Solving Repeatedly

Every situation in this exercise asks more than one question about the same relationship — the election problem asks about both 1200 voters and 800 total, the parking problem asks about three separate durations. Graphing once and reading multiple times is what makes that efficient: the alternative, re-solving the original equation from scratch for every single question, works just as correctly but throws away the obvious shortcut a straight line offers once it's already been drawn. The upfront cost of drawing the graph accurately is really an investment against every question that follows it — the first question asked about a given relationship might not obviously justify the effort of plotting, but the second and third almost always do, once it's clear more than one reading will be needed from the exact same line.

One More Shape of Linear Equation

Every equation graphed so far has involved both x and y together. Exercise 6.5 closes the chapter with the one remaining case — equations that mention only one variable, like x = 5 or y = −3 — and what their graphs turn out to look like once treated as genuine equations in two variables rather than one. It's a fitting close to the chapter: every equation graphed up to this point has tilted at some slope, rising or falling steadily as x increases, and the one remaining case — a line that doesn't tilt at all, running perfectly flat or perfectly upright instead — has been waiting quietly in the wings the entire time.