Algebra word problem: finding the distance

[NotaMathPerson]

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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!

 

[MarkFL]

I would begin this problem by drawing a diagram:

View attachment 5655

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

\(\displaystyle d=x+y+z\)

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

\(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=c+b-a\)

\(\displaystyle \frac{y}{a}+\frac{z}{b}+\frac{x}{c}=a+c-b\)

\(\displaystyle \frac{z}{a}+\frac{x}{b}+\frac{y}{c}=b+a-c\)

What do you get when you add these 3 equations?

 

[NotaMathPerson]

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

\(\displaystyle d=x+y+z\)

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

\(\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=c+b-a\)

\(\displaystyle \frac{y}{a}+\frac{z}{b}+\frac{x}{c}=a+c-b\)

\(\displaystyle \frac{z}{a}+\frac{x}{b}+\frac{y}{c}=b+a-c\)

What do you get when you add these 3 equations?

This is what I get

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

 

[MarkFL]

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:

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

Now, solve for $d$. :)

 

[NotaMathPerson]

Yes, good! (Yes)

I chose to write this as:

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

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.

 

[MarkFL]

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:

\(\displaystyle \frac{ax}{60}+\frac{by}{60}+\frac{cz}{60}=c+b-a\)

\(\displaystyle \frac{ay}{60}+\frac{bz}{60}+\frac{cx}{60}=a+c-b\)

\(\displaystyle \frac{az}{60}+\frac{bx}{60}+\frac{cy}{60}=b+a-c\)

Now when we add the equations, we obtain:

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