MHB Algebra word problem: finding the distance

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The discussion revolves around solving an algebra word problem to find the distance \(d\) using given speeds for walking, riding, and driving. The initial equations set up a relationship between the distances \(x\), \(y\), and \(z\) traveled at different speeds, leading to a combined equation that simplifies to \(d\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a+b+c\). A participant realizes that the velocities were mistakenly interpreted in hours instead of minutes, prompting a revised system of equations. Ultimately, by adding the equations correctly, they derive that the distance \(d\) equals 60 miles, confirming the answer from the book. The discussion emphasizes the importance of correctly interpreting the problem's parameters to arrive at the solution.
NotaMathPerson
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View attachment 5654

Hello! Please help continue solving tye problem I got stuck.

This is my attemptlet $d=$ length of the circuit
$\frac{60}{a}$mph --- speed for walking
$\frac{60}{b}$mph ----speed for riding
$\frac{60}{c}$mph ---- speed for driving

$d = \frac{60}{a}t_{1}+ \frac{60}{b}t_{2} + \frac{60}{c}t_{3}$

From here I cannot continue. Kindly help me. Thanks!
 

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I would begin this problem by drawing a diagram:

View attachment 5655

Now, what we want to find is the distance $d$ where:

$$d=x+y+z$$

Using the information given in the problem, we may write:

$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=c+b-a$$

$$\frac{y}{a}+\frac{z}{b}+\frac{x}{c}=a+c-b$$

$$\frac{z}{a}+\frac{x}{b}+\frac{y}{c}=b+a-c$$

What do you get when you add these 3 equations?
 

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MarkFL said:
I would begin this problem by drawing a diagram:
Now, what we want to find is the distance $d$ where:

$$d=x+y+z$$

Using the information given in the problem, we may write:

$$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=c+b-a$$

$$\frac{y}{a}+\frac{z}{b}+\frac{x}{c}=a+c-b$$

$$\frac{z}{a}+\frac{x}{b}+\frac{y}{c}=b+a-c$$

What do you get when you add these 3 equations?

This is what I get

$\left(x+y+z\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a+b+c$
 
NotaMathPerson said:
This is what I get

$\left(x+y+z\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a+b+c$

Yes, good! (Yes)

I chose to write this as:

$$d\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a+b+c$$

Now, solve for $d$. :)
 
MarkFL said:
Yes, good! (Yes)

I chose to write this as:

$$d\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=a+b+c$$

Now, solve for $d$. :)

I thought of it too. But solving for d only gives me letters. In my book the answer is 60 miles. Why is that?

And can you explain why do we need to add the 3 eqns? Thanks.
 
NotaMathPerson said:
I thought of it too. But solving for d only gives me letters. In my book the answer is 60 miles. Why is that?

And can you explain why do we need to add the 3 eqns? Thanks.

I didn't read the question thoroughly (regarding the 3 velocities being given in minutes instead of hours)...what we get instead is the system:

$$\frac{ax}{60}+\frac{by}{60}+\frac{cz}{60}=c+b-a$$

$$\frac{ay}{60}+\frac{bz}{60}+\frac{cx}{60}=a+c-b$$

$$\frac{az}{60}+\frac{bx}{60}+\frac{cy}{60}=b+a-c$$

Now when we add the equations, we obtain:

$$\frac{d}{60}(a+b+c)=a+b+c\implies d=60$$
 
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