A rational expression word problem

[hem]

I have been following greenemath.com to revise my algebra and in one of the solved examples, I have a problem of understanding.

The problem is in the screenshot and my difficulty is such that while Mary saves two hours when she follows the interstate, shouldn't the two hours be subtracted from x/50 rather than added to it?

 

[Harry_the_cat]

x/50 is the time taken on the interstate.

x/30 is the time taken on the highway.

Which one is the smaller? x/50 or x/30 ? This should help you understand why the 2 is added rather than subtracted.

 

[hem]

x/50 is the time taken on the interstate.

x/30 is the time taken on the highway.

Which one is the smaller? x/50 or x/30 ? This should help you understand why the 2 is added rather than subtracted.

Sorry, I have still not understood. X/50 is the smaller one because the speed is higher on the interstate than the highway which should logically reduce the time as mentioned in the problem by 2 hours then why is it added?

 

[Harry_the_cat]

OK think of it this way. Let's just say the time taken on the interstate is 6 hours. Then the time taken on the highway is 8 hours.
What have you got to do to 6 to get 8?

 

[hem]

OK think of it this way. Let's just say the time taken on the interstate is 6 hours. Then the time taken on the highway is 8 hours.
What have you got to do to 6 to get 8?

Add 2 hours

 

[Harry_the_cat]

Correct! So you need to add 2 hours to the interstate time to get the highway time.

 

[Harry_the_cat]

Correct! So you need to add 2 hours to the interstate time to get the highway time.

That is:

interstate time + 2 = highway time

x/50 + 2 = x/30

 

[Harry_the_cat]

Or you could write it as:

x/30 - 2 = x/50

Same thing! But I think that is how you are thinking it.

 

[hem]

Or you could write it as:

x/30 - 2 = x/50

Same thing! But I think that is how you are thinking it.

yes more or less. Thank you for your efforts, now the light is dawning.

 

[Harry_the_cat]

Happy to help!

 

[Dr.Peterson]

The problem is in the screenshot and my difficulty is such that while Mary saves two hours when she follows the interstate, shouldn't the two hours be subtracted from x/50 rather than added to it?

One thing I do in reading a word problem is to restate each fact as directly as possible. So when I read,

She saves two hours by driving on the interstate.​

I restate that (in my mind, or on paper) as

She takes 2 hours less on the interstate.​

Then I might go even further, and write

The time on the interstate is 2 hours less than the time on the highway.​

This can translate directly into either

x/50 = x/30 - 2​

or

x/50 + 2 = x/30​

Then I will check my thinking by observing that x/50 is necessarily less than x/30, so that this relationship makes sense.

They demonstrated similar thinking by restating the fact as

It takes 2 hours longer on the highway than on the interstate.​

I always check my thinking, because the wording in English is often a little twisted!

 

[JeffM]

I know this is an old post, but I believe it represents an important pedagogical point.

I believe we should teach beginning students of elementary algebra to assign a unique symbol to each unknown quantity, and then say we need as many equations as we have symbols to find a solution.

$$\text {distance in miles to state fair } = d,\\ \text {hours to make trip on interstate } = x, \text { and}\\ \text {hours to make trip on highway } = y.\\ x = \dfrac{d}{50}.\\ y = \dfrac{d}{30}.\\ x = y - 2 \text { or } y = x + 2.$$
Tracking the language of the word problem a la Dr. Peterson is good because it is highly intuitive even though formally irrelevant.

Now this approach demands that we teach a little formal algebra before actually trying to SOLVE word problems.

Substitution

$$x = \dfrac{d}{50} \implies y - 2 = \dfrac{d}{50}.$$
Clearing fractions.

$$y - 2 = \dfrac{d}{50} \implies 50(y - 2) = 50 \times \dfrac{d}{50} \implies 50y - 100.= d.\\ y = \dfrac{d}{30} \implies 30y = 30 \times \dfrac{d}{30} = d.$$
Creating equations for known equalities.

$$30y = d \text { and } 50y - 100 = d \implies 30y = 50y - 100 \implies\\ 30y - 30y = 50y - 30y - 100 \implies 0 = 20y - 100 \implies \\ 0 + 100 = 20y - 100 + 100 \implies 100 = 20y \implies 20y = 100 \implies\\ \dfrac{20y}{20} = \dfrac{100}{20} \implies y = 5.$$
But then it is obvious that

$$x = y - 2 = 5 - 2 = 3 \text { and }\\ 3 = \dfrac{d}{50} \implies 50 \times 3 = 50 \times \dfrac{d}{50} \implies 150 = d.\\ \text {CHECK:} \dfrac{150}{50}= 3 = x \text { as found before.}$$
I admit we do not teach algebra this way. I merely claim that we should, and if we did, this student would not have been confused by this problem.

 

[hem]

I know this is an old post, but I believe it represents an important pedagogical point.

I believe we should teach beginning students of elementary algebra to assign a unique symbol to each unknown quantity, and then say we need as many equations as we have symbols to find a solution.

$$\text {distance in miles to state fair } = d,\\ \text {hours to make trip on interstate } = x, \text { and}\\ \text {hours to make trip on highway } = y.\\ x = \dfrac{d}{50}.\\ y = \dfrac{d}{30}.\\ x = y - 2 \text { or } y = x + 2.$$
Tracking the language of the word problem a la Dr. Peterson is good because it is highly intuitive even though formally irrelevant.

Now this approach demands that we teach a little formal algebra before actually trying to SOLVE word problems.

Substitution

$$x = \dfrac{d}{50} \implies y - 2 = \dfrac{d}{50}.$$
Clearing fractions.

$$y - 2 = \dfrac{d}{50} \implies 50(y - 2) = 50 \times \dfrac{d}{50} \implies 50y - 100.= d.\\ y = \dfrac{d}{30} \implies 30y = 30 \times \dfrac{d}{30} = d.$$
Creating equations for known equalities.

$$30y = d \text { and } 50y - 100 = d \implies 30y = 50y - 100 \implies\\ 30y - 30y = 50y - 30y - 100 \implies 0 = 20y - 100 \implies \\ 0 + 100 = 20y - 100 + 100 \implies 100 = 20y \implies 20y = 100 \implies\\ \dfrac{20y}{20} = \dfrac{100}{20} \implies y = 5.$$
But then it is obvious that

$$x = y - 2 = 5 - 2 = 3 \text { and }\\ 3 = \dfrac{d}{50} \implies 50 \times 3 = 50 \times \dfrac{d}{50} \implies 150 = d.\\ \text {CHECK:} \dfrac{150}{50}= 3 = x \text { as found before.}$$
I admit we do not teach algebra this way. I merely claim that we should, and if we did, this student would not have been confused by this problem.

Thank you so much this explains the nuts and bolt