Finding Distance and Speed (Linear Equation Method)

Video Explanation

Question

A man walks a certain distance with a certain speed. If he walks \( \frac{1}{2} \) km/h faster, he takes 1 hour less. If he walks 1 km/h slower, he takes 3 hours more. Find the distance and original speed.

Solution

Step 1: Concept

Time = Distance / Speed

Step 2: Let Variables

Let original speed = \(x\) km/h

Let distance = \(d\) km

Step 3: Form Equations

\[ \frac{d}{x + \frac{1}{2}} = \frac{d}{x} – 1 \quad (1) \]

\[ \frac{d}{x – 1} = \frac{d}{x} + 3 \quad (2) \]

Step 4: Convert into Linear Form

Let:

\[ a = \frac{d}{x} \]

Then:

\[ \frac{d}{x + \frac{1}{2}} = \frac{ax}{x + \frac{1}{2}}, \quad \frac{d}{x – 1} = \frac{ax}{x – 1} \]

— From (1):

\[ \frac{ax}{x + \frac{1}{2}} = a – 1 \]

Multiply:

\[ ax = (a – 1)\left(x + \frac{1}{2}\right) \]

\[ ax = ax + \frac{a}{2} – x – \frac{1}{2} \]

Cancel \(ax\):

\[ 0 = \frac{a}{2} – x – \frac{1}{2} \]

\[ a – 2x = 1 \quad (3) \]

— From (2):

\[ \frac{ax}{x – 1} = a + 3 \]

Multiply:

\[ ax = (a + 3)(x – 1) \]

\[ ax = ax – a + 3x – 3 \]

Cancel \(ax\):

\[ 0 = -a + 3x – 3 \]

\[ a = 3x – 3 \quad (4) \]

Step 5: Solve Linear Equations

From (3):

\[ a = 2x + 1 \]

From (4):

\[ a = 3x – 3 \]

Equate:

\[ 2x + 1 = 3x – 3 \]

\[ x = 4 \]

Step 6: Find Distance

\[ a = \frac{d}{x} \Rightarrow a = 2x + 1 = 9 \]

\[ d = ax = 9 \times 4 = 36 \]

Conclusion

\[ \text{Speed} = 4 \text{ km/h}, \quad \text{Distance} = 36 \text{ km} \]

Verification

Faster speed = 4.5 km/h → time = 8 hrs

Original = 9 hrs → 1 hour less ✔

Slower speed = 3 km/h → time = 12 hrs

Original = 9 hrs → 3 hours more ✔