Linear Equations in Real Situations

Linear equations are not just exam material , they describe a remarkable slice of real-world behaviour. The cost of a taxi ride. The distance covered at a constant speed. The temperature converted from Celsius to Fahrenheit. A loan repayment plan with fixed monthly instalments. This lesson shows how to set up the equation, draw its graph, and read answers off the graph.

Definitions

A linear model is a situation in which two quantities change in such a way that their relationship can be written as a linear equation $ax + by + c = 0$ . The defining feature is proportionality with at most a fixed offset: changing one variable by a unit changes the other by a fixed amount, regardless of where you started.

Concept and four classic linear models

Model 1: Total cost with a fixed and variable component. Suppose a taxi charges a flag-down fare of $\text{Rs.}~25$ plus $\text{Rs.}~15$ per kilometre. If $x$ is the distance in km and $y$ the total fare in Rs., $y = 25 + 15 x.$ This is linear in $x$ and $y$ , and its graph is a straight line with $y$ -intercept $25$ (cost when $x = 0$ ) and slope $15$ (extra cost per extra km).

Model 2: Two-quantity budget. A shopper buys notebooks at $\text{Rs.}~30$ each and pens at $\text{Rs.}~20$ each, spending exactly $\text{Rs.}~600$ . If $x$ = number of notebooks and $y$ = number of pens, $30 x + 20 y = 600 \iff 3x + 2y = 60.$ Setting up the equation is the whole task; finding integer solutions $(x, y)$ that make sense (both non-negative whole numbers) is the second step.

Model 3: Distance at constant speed. A train travels at $60$ km/h. If $x$ is time in hours and $y$ is distance in km, $y = 60 x.$ This is a linear equation through the origin: at $x = 0$ (just started), $y = 0$ (no distance yet); at $x = 2$ , $y = 120$ km.

Model 4: Temperature conversion. The Celsius-to-Fahrenheit conversion is $F = \frac{9}{5} C + 32.$ This is a linear equation in $C$ and $F$ . At $C = 0$ , $F = 32$ (water freezes); at $C = 100$ , $F = 212$ (water boils). The graph is a straight line passing through $(0, 32)$ and $(100, 212)$ .

The general procedure for word problems.

Identify the two varying quantities. Call them $x$ and $y$ (or use letters that match the context, like $C$ and $F$ ).
Spot the constants in the situation. Per-unit rates, fixed fees, initial values.
Write the equation using the relation given by the problem.
Simplify to standard form if asked, or to $y = mx + c$ form if you want to graph it quickly.
Generate a table or two intercepts and draw the graph.
Use the graph to answer specific questions like "what is $y$ when $x = 3$ ?" or "for which $x$ is $y > 100$ ?"

Reading domain constraints from context. In real situations, $x$ and $y$ often have natural restrictions. The number of pens you buy is a non-negative integer. The distance travelled is non-negative. Time elapsed is non-negative. The graph is then drawn on the relevant portion of the plane, not the whole plane. We still call the equation linear; we just restrict its domain.

An interpretive moment. In a linear model, the $y$ -intercept usually carries a real-world meaning: it is the value of $y$ when $x = 0$ . The slope tells you the rate of change. So in the taxi example, the $y$ -intercept $25$ is the cost at the moment of getting in (before moving), and the slope $15$ is the cost per km. Numbers from the equation acquire concrete meanings.

Worked examples

Example 1. A book costs $\text{Rs.}~50$ plus $\text{Rs.}~5$ per page. Write a linear equation and find the cost of a $40$ -page book.

Let $x$ = number of pages, $y$ = cost. Equation: $y = 50 + 5x$ . At $x = 40$ : $y = 50 + 200 = 250$ . Cost: $\text{Rs.}~250$ .

Example 2. The work charge for an electrician is $\text{Rs.}~100$ plus $\text{Rs.}~50$ per hour. Write the equation and find the charge for $3$ hours of work.

Let $x$ = hours, $y$ = charge. $y = 100 + 50x$ . At $x = 3$ : $y = 250$ . Rs. $250$ .

Example 3. Sara has $\text{Rs.}~100$ to spend. Apples cost $\text{Rs.}~10$ each ( $x$ apples) and oranges $\text{Rs.}~5$ each ( $y$ oranges). She spends all her money. Write the equation.

$10 x + 5 y = 100 \Rightarrow 2x + y = 20$ . Integer solutions: $(0, 20), (1, 18), \ldots, (10, 0)$ .

Example 4. A taxi charges $\text{Rs.}~25$ for the first km and $\text{Rs.}~10$ per km thereafter. If $x \ge 1$ is the total distance in km, write the cost equation.

Cost $y = 25 + 10(x - 1) = 15 + 10x$ for $x \ge 1$ . Linear in $x, y$ .

Example 5. Convert $25^\circ$ C to Fahrenheit using $F = \dfrac{9}{5} C + 32$ .

$F = \dfrac{9 \cdot 25}{5} + 32 = 45 + 32 = 77$ . So $25^\circ$ C $= 77^\circ$ F.

Try it yourself

A pen costs $\text{Rs.}~x$ and a notebook costs $\text{Rs.}~y$ . Two pens and three notebooks cost $\text{Rs.}~80$ . Write the equation.
A car travels at $40$ km/h. Write a linear equation between time $t$ (hours) and distance $d$ (km). Graph the line.
The cost of $3$ apples and $2$ oranges is $\text{Rs.}~70$ . Write a linear equation in two variables with these prices.
Anil earns $\text{Rs.}~300$ a day plus a bonus of $\text{Rs.}~20$ per item sold. Write an equation for total daily earnings $y$ in terms of items sold $x$ .
Convert $40^\circ$ C to Fahrenheit.
For which Celsius temperature does $C = F$ ? Hint: solve $C = \tfrac{9}{5} C + 32$ .
A taxi charges $\text{Rs.}~30$ plus $\text{Rs.}~12$ per km. Find the fare for $7$ km and write a linear equation.
The cost of $4$ chairs and $2$ tables is $\text{Rs.}~700$ , and the cost of $1$ chair and $3$ tables is $\text{Rs.}~600$ . Write the two linear equations (don't solve; that's for next year).
Plot the graph of $y = \tfrac{9}{5} x + 32$ on a Celsius–Fahrenheit grid. Mark the temperatures $0, 25, 100$ Celsius.
The total amount in a piggy bank after $n$ weeks is $A = 50 + 20n$ . Find $A$ after $10$ weeks, and find $n$ when $A = 250$ .

Pitfalls / Insight

Use letters that suit the context. It is fine to write $F = \tfrac{9}{5} C + 32$ instead of forcing $x$ and $y$ .
Mind domain restrictions. Number of items must be a non-negative integer. Time and distance are non-negative. Plot only the relevant portion.
The $y$ -intercept is the "starting value". The slope is the rate of change. Both have direct real-world meaning in a linear model.

Insight. Almost any situation with a "fixed cost plus per-unit cost" or "constant speed" or "linear conversion" structure is a linear equation in disguise. Once you spot the pattern, you can set up the equation in two seconds, draw the graph in two minutes, and answer any quantitative question from the picture. That is the practical payoff of this short chapter.