Problems-solving in Equations The distance between the two cities A and B is 300 km, set off a car from the city a toward the city b by speed 90 km/h and set off from the city b bicycle toward a by speed of 10km/h.  if you knew that the car and the bike were based in nine in the morning. Select the time that the car and the bike are going to met ?      
Equation for the 8th grade ! 
Please help i need to get a formulation of this equation 
 A: Hint: 
$$ v = \frac{\mathrm{d} x}{\mathrm{d} t} = 100 $$
Integrating, you get a very simple equation of x. If you set x(0) = 0, the equation simplifies to remove the constant. 
Now, x = 300. Solve for t. 
A: To find the answer you are looking for in a more visual way, consider this diagram.

After 1 hour, the car (red line) traveled 90km from city A, and the bike (green line) traveled 10km from city B. Now, draw another red line that is 90 km long, right after the red one. It will go from 90km to 180km. Then, draw another green line that is 10km long, right after the green one. It will go from 290km to 280km. Having two lines will tell you how far they traveled after 2 hours. Keep this up until the red and green lines meet. 
A: Here are two ways of tackling this:
1)
Let $e_1$ be the distance traveled by the car in $t_1$ hours.
Let $e_2$ be the distance traveled by the bike in $t_2$ hours.
When the car and the bike meet, the will have been traveling for the same amount of time (since they set off at the the same time). Then, $t_1=t_2$.
$$e_1+e_2=300km$$
$$v_{car} \, t_1 +  v_{bike} \, t_2 = 300km$$
$$90\frac{km}{h} \, t_1 +10\frac{km}{h} \, t_1  = 300km$$
$$100\frac{km}{h} \, t_1 =300km$$
$$t_1=3h$$
2) Notice that the distance between the car and the bike gets shortened by $100 \frac{km}{h}$ and you get again $100 \frac{km}{h} \, t = 300km$.
If they set off at nine, they will meet at twelve.
A: Let $d_{car}$ the distance that the car will run until it meets the bike, $d_{bike}$ the distance that the bike will run until it meets the car.
We get $d_{car}+d_{bike}=AB=300km$.
But we have $d=v\times t$ where $d$ is the distance, $v$ the velocity (speed), and $t$ the time.
Can you continue like this?
