Direct variation

Two quantities $x$ and $y$ are said to vary directly — or be in direct proportion — if, as one changes, the other changes in exactly the same ratio. Symbolically,

where: $x$ and $y$ are the two varying quantities, and $k$ is a fixed number called the constant of variation (it does not change as $x$ and $y$ change). The equation $y = kx$ says $y$ is always $k$ times $x$ — double $x$ and $y$ doubles, triple $x$ and $y$ triples, halve $x$ and $y$ halves. A quick test for direct variation: tabulate a few pairs of values and compute $y/x$ for each. If the ratio comes out the same every time, the quantities vary directly; if it keeps changing, they do not. This is worth doing before assuming a word problem is a direct variation at all. For any two corresponding pairs of values $(x_1, y_1)$ and $(x_2, y_2)$ of the same two varying quantities,

where: the subscripts $1$ and $2$ label two different snapshots of the same relationship — for example, the cost and quantity on Monday versus the cost and quantity on Tuesday, at the same fixed rate. Cross-multiplying gives the equivalent proportion form

which is often faster than finding $k$ explicitly when you only need one unknown value. Many students solve the very same problems using the unitary method instead: find the value of $y$ corresponding to one unit of $x$, then scale that single-unit value up or down to whatever amount the question asks for. The unitary method and the $k$-based method are the same idea in different clothes — finding $y$ for one unit of $x$ is finding $k$ — so both must give the same answer, and checking one against the other is a good habit against careless slips. The picture below shows why the graph of a direct variation is always a straight line through the origin: because $y = kx$, every point $(x, y)$ that satisfies the relation lies on a line of slope $k$ passing through $(0,0)$ — no notebooks bought means no money spent, no time travelled means no distance covered.

Direct variation is everywhere outside the classroom too: the bill you pay is directly proportional to the quantity bought at a fixed rate, distance covered is directly proportional to time at a constant speed, and the weight of a metal rod is directly proportional to its length as long as its thickness stays the same. Spotting the phrase "at a constant rate/price/speed" in a word problem is usually the clue that direct variation is the right tool.

The cost of 8 identical notebooks is ₹296. Find the cost of 13 such notebooks.

Step 1. Identify the variation. Cost varies directly with the number of notebooks bought at a fixed price, so $\dfrac{\text{Cost}}{\text{Number}} = k$, with $x_1 = 8$ notebooks and $y_1 = ₹296$.

Step 2. Find $k$, the cost of one notebook.

So each notebook costs ₹37.

Step 3. Use direct variation to find the cost of 13 notebooks.

Step 4. Check with the proportion form. With $x_1 = 8,\ y_1 = 296,\ x_2 = 13,\ y_2 = ?$,

Both methods agree. 13 notebooks cost ₹481.

Competitive stretch

A car travelling at a steady speed covers 165 km in 3 hours. How long will it take to cover 385 km at the same speed? Here distance $d$ varies directly with time $t$: $d = kt$, so $k = \dfrac{165}{3} = 55$ km/h. Setting $385 = 55t$ gives $t = \dfrac{385}{55} = 7$. A proportion check, $\dfrac{165}{3} = \dfrac{385}{t}$, gives $165t = 1155$, so $t = 7$ again. The car takes 7 hours. Notice the same four-step method solved a completely different real-world pair — cost/notebooks and distance/time — because both situations are, underneath, the same equation $y = kx$.

• Confusing direct with inverse variation. In direct variation, both quantities increase or decrease together; do not cross-multiply as if one increasing meant the other decreasing — that pattern belongs to inverse variation instead.
• Assuming direct variation without checking. Not every "more of one, more of the other" situation is a clean direct variation. Verify with at least two pairs of values that $y/x$ is genuinely the same constant before applying $y = kx$.
• Setting up the proportion upside down. Writing $\dfrac{x_1}{y_1} = \dfrac{y_2}{x_2}$ instead of $\dfrac{x_1}{y_1} = \dfrac{x_2}{y_2}$ gives a wrong, often absurd, answer — keep the same quantity in the numerator on both sides.
• Mixing units mid-problem, such as leaving one distance in metres and another in kilometres, which silently corrupts the constant $k$.

Ask the tutor: If you have a word problem and are not sure whether it is direct variation, share it with the AI tutor and ask it to identify the constant-rate phrase that gives it away.