---
title: "Fit y = ax/(b + x)"
subject: "Numerical Method"
chapter: 4
key: "semester_4/numerical_method:fit-y-ax-over-b-plus-x"
source: https://www.bctengineer.com/q/semester_4/numerical_method/fit-y-ax-over-b-plus-x
site: BCT Engineering
---

# Fit y = ax/(b + x)

Numerical Method · Chapter 4: Interpolation

- Asked in: 2068 Magh, 8 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/fit-y-ax-over-b-plus-x
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-4

## Question

Fit the following set of data into a curve y = ax/(b + x).
| X | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Y | 0.500 | 0.667 | 0.750 | 0.2 | 0.833 |

## Short answer

Linearize to 1/y = 1/a + (b/a)(1/x), compute normal equations, yielding a = 0.9999, b = 0.9995. Thus the fitted curve is y = 0.9999x/(0.9995 + x), essentially y = x/(1+x).

## Answer

**Assumption.** The printed y at x = 4 is 0.2, which breaks the pattern of the other values (0.500, 0.667, 0.750, … 0.833, all of the form x/(1+x)); it is taken as a misprint of 0.800.

#### Linearisation

y = ax/(b + x)  →  1/y = (b + x)/(ax) = 1/a + (b/a)(1/x)

Put Y = 1/y, X = 1/x, then Y = A + BX, with A = 1/a and B = b/a.

| x | y | X = 1/x | Y = 1/y | X² | XY |
|---|---|---|---|---|---|
| 1 | 0.500 | 1.0000 | 2.0000 | 1.0000 | 2.0000 |
| 2 | 0.667 | 0.5000 | 1.4993 | 0.2500 | 0.7496 |
| 3 | 0.750 | 0.3333 | 1.3333 | 0.1111 | 0.4444 |
| 4 | 0.800 | 0.2500 | 1.2500 | 0.0625 | 0.3125 |
| 5 | 0.833 | 0.2000 | 1.2005 | 0.0400 | 0.2401 |
| Σ |  | 2.2833 | 7.2831 | 1.4636 | 3.7467 |

#### Normal equations

```text
ΣY = n·A + B·ΣX      →  7.2831 = 5A + 2.2833B
ΣXY = A·ΣX + B·ΣX²   →  3.7467 = 2.2833A + 1.4636B
```

Solving: B = (nΣXY − ΣXΣY)/(nΣX² − (ΣX)²) = 0.9996,  A = (ΣY − BΣX)/n = 1.0001.

Hence a = 1/A = 0.9999 and b = B/A = 0.9995.

**Answer: y = 0.9999x/(0.9995 + x)**, which is practically y = x/(1 + x). Check: x = 3 gives 0.750 against 0.750.

If the printed value 0.2 is used literally, the same method gives a ≈ 0.386, b ≈ −0.325, a poor and meaningless fit (negative b), which supports the misprint reading.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is y = ax/(b + x) linearised?
   A. y = a + bx
   B. y² = ax
   C. log y = log a + b log x
   D. 1/y = 1/a + (b/a)(1/x)
   Answer: D. Taking reciprocals gives a linear relation in 1/x. (Key point: Linearisation)

2. If Y = A + BX, how are a and b found?
   A. a = A, b = B
   B. a = 1/A, b = B/A
   C. a = 1/B, b = A
   D. a = AB
   Answer: B. A = 1/a and B = b/a. (Key point: Constants)

3. Why is y = 0.2 at x = 4 treated as a misprint?
   A. x = 4 is missing
   B. It is negative
   C. It breaks the x/(1+x) pattern; 0.800 fits
   D. It is too large
   Answer: C. Literal use gives meaningless negative b. (Key point: Misprint)

## Close variants set in other exams

- [Fit curve y = a b^x (decay data)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Afit-y-abx-d) — Numerical Method, chapter 4, asked 3×

## Related in Numerical Method

- [Fit y = ln(ax + b)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Afit-log-ax-b) — Numerical Method, chapter 4, asked 2×
- [Fit curve y = a x^b (A)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Afit-y-ax-b) — Numerical Method, chapter 4, asked 2×
- [Fit y = a b^x (growth data)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Afit-y-abx-d2) — Numerical Method, chapter 4
- [Fit y = a b^x and find y(7)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Afit-y-abx-y7) — Numerical Method, chapter 4, asked 2×
- [Fit y = a e^(bx)](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Afit-y-aebx) — Numerical Method, chapter 4
- [Least squares: y = ax + b + c/x](https://www.bctengineer.com/question.md?key=semester_4%2Fnumerical_method%3Aleast-squares-ax-b-c-over-x) — Numerical Method, chapter 4

## Related in other subjects

- [Vertical curve +0.5% and -0.7%, IP RL 350.750](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-pos05-neg07-ip-500) — Surveying II, chapter 6
- [Summit curve limited to 750 m](https://www.bctengineer.com/question.md?key=semester_6%2Ftransportation_engineering_i%3An-75bh-vc-summit-750) — Transportation Engineering I, chapter 3, asked 3×
- [Vertical curve -1% and +0.5%, IP 500 m](https://www.bctengineer.com/question.md?key=semester_4%2Fsurveying_ii%3Avertical-curve-falling-1-rising-05) — Surveying II, chapter 6, asked 2×
- [Least squares fit for calibration data](https://www.bctengineer.com/question.md?key=semester_3%2Fmetrology%3Aleast-squares-calibration-line) — Metrology, chapter 2