.
Let " t " be the time for old boiler to burn the load coal (in "days").
Then the time for new boiler to burn the load coal will be (t+5) days.
Thus the rate of burning for the old boiler is of the total load per day;
the rate of burning for the new boiler is of the total load per day.
When the two boilers work simultaneously, their combined rate is the sum + ,
and it is equal to , according to the condition.
So, your equation is
+ = .
To solve it, first multiply both sides by 6t*(t+5); then simplify
6*(t+5) + 6t = t*(t+5)
6t + 30 + 6t = t^2 +5t
t^2 - 7t -30 = 0
(t-10)*(t+3) = 0
Only positive root t = 10 days is meaningful.
Answer. 10 days for old boiler and 15 days for new boiler.
CHECK. + = = = . ! Correct !
Solved.
------------------
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. See the lesson
- Using quadratic equations to solve word problems on joint work
Read it and get be trained in solving joint-work problems.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lesson is the part of this textbook under the topic
"Rate of work and joint work problems" of the section "Word problems".
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.