# 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 !

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Where are the bike and car after 1 hour? – John Habert Feb 27 '14 at 15:37
Hint: they move towards each other at 100km/h – user130512 Feb 27 '14 at 15:39
well , the car is moving at 90km/h and the bike at 10km/h . i didn't really get it how did you make it 100km/h and why ? Sorry but i'm really not good in solving-problems in equations i'm only good at equation and others – Manal Feb 27 '14 at 15:44

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.

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It's true yes but i need an equation here in term of x – Manal Feb 27 '14 at 16:19
Well if that's the case, let $x$ be how many hours the car and bike have traveled. Then, translating the diagram to an equation, we want to find when: $90x + 10x = 300$. In other words, how many reds and greens does it take to add up to 300 (cover the entire number line). Solving for $x$ will give you how many hours are needed to make that happen. – josh Feb 27 '14 at 16:53
Thanks aloot yes it's the answer i've been looking for ! – Manal Feb 27 '14 at 16:56

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.

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Where is the formulation of the equation ? Sorry but didn't understand – Manal Feb 27 '14 at 16:14
The OP was looking for an 8th grade solution ... – Michael Hoppe Feb 27 '14 at 16:56
I'm pretty sure he didn't originally indicate he was 8th grade, but my bad (just checked his edit history, and I was correct). – Chantry Cargill Feb 27 '14 at 17:37

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.

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Yes it's right ! but i needed the answer in terms of x and with one unkown – Manal Feb 27 '14 at 16:58

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?

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Sorry , yeah i can but i already found the answer ! thanks for help – Manal Feb 27 '14 at 17:02
No problem, glad you found the answer :) – zagoku Feb 27 '14 at 17:06