Solve the following linear simultaneous equations by elimination method: 6x+y=15, 3x+y=9

Let $D$ be the total distance of the journey in km.
For the first part of the journey:
The distance covered is $D_1 = \frac{2}{5}D$.
The speed is $S_1 = 30 \text{ km/h}$.
The time taken for the first part is $T_1 = \frac{D_1}{S_1} = \frac{\frac{2}{5}D}{30} = \frac{2D}{150} = \frac{D}{75}$ hours.

For the second part of the journey (remaining distance):
The remaining distance is $D_2 = D - D_1 = D - \frac{2}{5}D = \frac{3}{5}D$.
The speed is $S_2 = 40 \text{ km/h}$.
The time taken for the second part is $T_2 = \frac{D_2}{S_2} = \frac{\frac{3}{5}D}{40} = \frac{3D}{200}$ hours.

The total time for the whole journey is given as 4 hours.
So, $T_1 + T_2 = 4$.

$$\frac{D}{75} + \frac{3D}{200} = 4$$

To solve for $D$, find the least common multiple (LCM) of the denominators 75 and 200.
$75 = 3 \times 5^2$
$200 = 2^3 \times 5^2$
The LCM$(75, 200) = 2^3 \times 3 \times 5^2 = 8 \times 3 \times 25 = 600$.
Multiply the entire equation by 600 to eliminate the denominators:

$$600 \left( \frac{D}{75} \right) + 600 \left( \frac{3D}{200} \right) = 4 \times 600$$ $$8D + 3(3D) = 2400$$ $$8D + 9D = 2400$$ $$17D = 2400$$ $$D = \frac{2400}{17} \text{ km}$$

Perform the division $2400 \div 17$:
$2400 \div 17 = 141$ with a remainder of 3.
So, $D = 141 + \frac{3}{17}$.
Now, convert the fraction $\frac{3}{17}$ to a repeating decimal:

$$\frac{3}{17} = 0.\overline{1764705882352941}$$

The repeating block consists of 16 digits.
Therefore, the total distance $D$ is:

$$D = 141.\overline{1764705882352941} \text{ km}$$

The total distance of the journey is $\text{141.}\overline{\text{1764705882352941}} \text{ km}$.