Class X Mathematics Worksheet on Pair of Linear Equations in Two Variables

Problem:

A man walks a certain distance with a certain speed. If he walks 3 km/h faster, he takes 1 hour less. But if he walks 3 km/h slower, he takes 3 hours more. Find the distance covered by the man and his original speed.

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Solution:

Let us denote:
- The original speed of the man as \( v \) (in km/h).
- The distance covered by the man as \( d \) (in km).

The time taken to cover the distance at the original speed is given by:
\[
t = \frac{d}{v}
\]

#### Case 1: Walking 3 km/h faster
If the man walks 3 km/h faster, his speed becomes \( v + 3 \). The time taken to cover the same distance is:
\[
t_1 = \frac{d}{v + 3}
\]
According to the problem, this time is 1 hour less than the original time:
\[
\frac{d}{v + 3} = \frac{d}{v} - 1
\]

#### Case 2: Walking 3 km/h slower
If the man walks 3 km/h slower, his speed becomes \( v - 3 \). The time taken to cover the same distance is:
\[
t_2 = \frac{d}{v - 3}
\]
According to the problem, this time is 3 hours more than the original time:
\[
\frac{d}{v - 3} = \frac{d}{v} + 3
\]

We now have two equations:
1. \(\frac{d}{v + 3} = \frac{d}{v} - 1\)
2. \(\frac{d}{v - 3} = \frac{d}{v} + 3\)

#### Step 1: Solve the first equation
Start with:
\[
\frac{d}{v + 3} = \frac{d}{v} - 1
\]
Multiply through by \( v(v + 3) \) to clear the denominators:
\[
d \cdot v = d \cdot (v + 3) - v(v + 3)
\]
Simplify:
\[
dv = dv + 3d - v^2 - 3v
\]
Subtract \( dv \) from both sides:
\[
0 = 3d - v^2 - 3v
\]
Rearrange:
\[
v^2 + 3v - 3d = 0 \quad \text{(Equation 3)}
\]

#### Step 2: Solve the second equation
Start with:
\[
\frac{d}{v - 3} = \frac{d}{v} + 3
\]
Multiply through by \( v(v - 3) \) to clear the denominators:
\[
d \cdot v = d \cdot (v - 3) + 3v(v - 3)
\]
Simplify:
\[
dv = dv - 3d + 3v^2 - 9v
\]
Subtract \( dv \) from both sides:
\[
0 = -3d + 3v^2 - 9v
\]
Rearrange:
\[
3v^2 - 9v - 3d = 0
\]
Divide through by 3:
\[
v^2 - 3v - d = 0 \quad \text{(Equation 4)}
\]

#### Step 3: Solve the system of equations
We now have two quadratic equations in \( v \) and \( d \):
1. \( v^2 + 3v - 3d = 0 \)
2. \( v^2 - 3v - d = 0 \)

Subtract Equation 4 from Equation 3:
\[
(v^2 + 3v - 3d) - (v^2 - 3v - d) = 0
\]
Simplify:
\[
v^2 + 3v - 3d - v^2 + 3v + d = 0
\]
\[
6v - 2d = 0
\]
Solve for \( d \):
\[
2d = 6v
\]
\[
d = 3v \quad \text{(Equation 5)}
\]

#### Step 4: Substitute \( d = 3v \) into one of the original equations
Substitute \( d = 3v \) into Equation 4:
\[
v^2 - 3v - d = 0
\]
\[
v^2 - 3v - 3v = 0
\]
\[
v^2 - 6v = 0
\]
Factorize:
\[
v(v - 6) = 0
\]
So, \( v = 0 \) or \( v = 6 \). Since \( v = 0 \) is not feasible (speed cannot be zero), we have:
\[
v = 6
\]

#### Step 5: Find \( d \)
Using \( d = 3v \):
\[
d = 3 \cdot 6 = 18
\]

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Final Answer:

The distance covered by the man is \( 18 \) km, and his original speed is \( 6 \) km/h.

\[
\boxed{18 \text{ km}, 6 \text{ km/h}}
\]