Question 1131263
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I deleted my first version and re-folmulated my post in this way (as follows):



            The solution depends on how you  (or me;  or another reader)  interpret this passage  "three times faster".


            In English,  it is a "dark form",  and native English speakers do not use it; 
            at least,  they do not use it when formulate Math problems.


            The correct form in English is to say  "One roofer's rate of work is  3  times as that of another".


            So,  I will re-formulate your post in this way:



<pre>
    Working together, two roofers can put a new roof in 5 days.  One roofer's  rate of work is three times that of another. 
    How long would it take the faster roofer to complete the job working alone?
</pre>


Then the solution is as follows.


<pre>
From the condition, you can write this system of 2 equations in 2 unknowns


{{{5/A}}} + {{{5/B}}} = 1         (1)

B = 3A,             (2)


where A is the time of the faster roofer and B is the time of the slower roofer.


Substitute (2) into equation (1). You will get

{{{5/A}}} + {{{5/(3A)}}} = 1


Multiply by 3A both sides.  You will get


15 + 5 = 3A

3A = 20

A = {{{20/3}}} hours = {{{6}}}{{{2/3}}} days is the time for the faster roofer.


Time for the slower roofer is 3 times this, i.e.  20 days.
</pre>

Solved.


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It is a standard and typical joint work problem.


There is a wide variety of similar solved joint-work problems with detailed explanations in this site. &nbsp;See the lessons

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Word-problems-WORKING-TOGETHER-by-Fractions.lesson>Using Fractions to solve word problems on joint work</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=http://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Solving-more-complicated-word-problems-on-joint-work.lesson>Solving more complicated word problems on joint work</A> 

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Rate-of-work-word-problems/Selected-problems-from-the-archive-on-joint-work-word-problems.lesson>Selected joint-work word problems from the archive</A> 



Read them and get be trained in solving joint-work problems.


Also, &nbsp;you have this free of charge online textbook in ALGEBRA-I in this site

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>.


The referred lessons are the part of this textbook under the topic 
"<U>Rate of work and joint work problems</U>" &nbsp;of the section &nbsp;"<U>Word problems</U>".



Save the link to this online textbook together with its description


Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson


to your archive and use it when it is needed.


=================


In your post, you insinuate the problem would be solved algebraically with the use of system of 2 equations.


There is another way of mental solution without using equation/equations.


<pre>
    Under given condition, two roofers are equivalent to 4 slower roofers.

    So, we are given that 4 slower roofers can complete the job in 5 days, working together.

    It implies immediately, that one single slower roofer can complete the job in 20 = 5*4 days.

    Then the time for one faster roofer to do it alone is  {{{20/3}}} = {{{6}}}{{{2/3}}} days - the same answer.
</pre>