1. ## Distance word problem

I am having trouble with this one. I am more interested in the steps than the answer.

A trip from Here to There is 2108 miles. If one car drove 6mph faster than a truck and got There 3 hours before the truck, find the speeds of the car and truck.

I think the car's speed is (truck's speed) + 6, its time is (Truck's time) - 3, and its distance is 2108, but I am unable to get an answer. I have manipulated the d=rt formula but to no avail. Any help would be appreciated.

2. Originally Posted by aliveiam
I am having trouble with this one. I am more interested in the steps than the answer.

A trip from Here to There is 2108 miles. If one car drove 6mph faster than a truck and got There 3 hours before the truck, find the speeds of the car and truck.

I think the car's speed is (truck's speed) + 6, its time is (Truck's time) - 3, and its distance is 2108, but I am unable to get an answer. I have manipulated the d=rt formula but to no avail. Any help would be appreciated.
Hi aliveiam,

Let's see....There's probably an easier way to approach this, but this is what I saw first.

Let r = the truck's rate

Let t = the truck's time

Let r + 6 = the car's rate
Let t - 3 = the car's time (the car gets there 3 hours ahead of the truck)

D = 2108 for both of them

Using D = rt,

[1] ${\color{red}rt=2108}$

[2] ${\color{blue}(r+6)(t-3)=2108}$

Solve the system above.

3. $\begin{cases}
Vt=2108 \\
(V+6)(t-3)=2108
\end{cases}$

Find V (truck's speed) and t (truck's time) out of it. Answers will be: truck's speed is 62, car's speed is 68.

4. ## I am still having trouble

I appreciate your time with this problem, but I am still unable to come up with the correct solutions. When I begin with [r + 6][t - 3] = 2108, I want to solve for r so I change t in terms of r and I get [r + 6][(distance/rate) - 3] = 2108 which becomes [r + 6][(2108/(r + 6)) - 3] = 2108. When I solve for r, however, I get -6. Do you know where I went wrong? Thank you for your time.

5. You're trying to solve it, using only one equation. Take t (2108/r) from the first one and put it in the second instead of finding t in the second one and putting it back there.

6. Originally Posted by aliveiam
I appreciate your time with this problem, but I am still unable to come up with the correct solutions. When I begin with [r + 6][t - 3] = 2108, I want to solve for r so I change t in terms of r and I get [r + 6][(distance/rate) - 3] = 2108 which becomes [r + 6][(2108/(r + 6)) - 3] = 2108. When I solve for r, however, I get -6. Do you know where I went wrong? Thank you for your time.
Let me finish it up for you.

Originally Posted by masters
Hi aliveiam,

Let's see....There's probably an easier way to approach this, but this is what I saw first.

Let r = the truck's rate

Let t = the truck's time

Let r + 6 = the car's rate
Let t - 3 = the car's time (the car gets there 3 hours ahead of the truck)

D = 2108 for both of them

Using D = rt,

[1] ${\color{red}rt=2108}$

[2] ${\color{blue}(r+6)(t-3)=2108}$

Solve the system above.
[2] ${\color{blue}(r+6)(t-3)=2108}$

$(r+6)(r-3)=2108$

$rt+6t-3r-18=2108$

Remember: [2] ${\color{red}rt=2108}$

$2108+6t-3r-18=2108$

[3] $6t-3r=18$

Now, use [1] ${\color{red}rt=2108}$ to solve for t in terms of r.

$t=\frac{2108}{r}$

Replace t in [3] $6t-3r=18$

$6\left(\frac{2108}{r}\right)-3r=18$

$\frac{12648}{r}-3r=18$

Multiply every term by r and rearrange to:

$3r^2+18r-12648=0$

Divide every term by 3.

$r^2+6r-4216=0$

Factor or use the quadratic formula:

$(r-62)(r+68)=0$

$r=62$

Ignore the negative root as it is extraneous (can't have a negative rate)

So the truck goes 62 mph and the car goes 62 + 6 = 68 mph.

7. ## Thank you much

I see where I went wrong. Thank you for your support. All is well.