One year on Mercury is equal to 87.97 Earth days. One year on Pluto is three times the length of one Mercury year minus 16.21 days. How long is one year on Pluto?

Sorry it is a bit long but I wished to explain about ambiguities in the question and the derivation of units/equations. The actual calculations are short! With assumptions I get ## ~~ 0.69color(white)(.) “Earth years”##
This is a tricky one as there may be some ambiguity about 16.21 days which is: to which planet is the day attributed? Also the units are tricky. They behave the same way numbers do!!!
##color(blue)(“Assumption 1”)##
From the sentence part “of one Mercury year minus 16.21 days” I am assuming that the days are Mercury days. From year minus 16.21 days” I am assuming that they are directly linked. Thus the days are directly attributable to the Mercury year.
##color(blue)(“Assumption 2”)##
Our annual cycle is split up into 365 units of cycles in one solar year ( a day). The other planets will circle the sun at different speeds but will experience the equivalent effects our planet does. However, these will be at different rates. Thus for each planet their year may also be split up into 365 rotational units.
##color(blue)(“Building the initial equation”)##
We are talking about units of measurement in both days and years. So let the generic unit for day be d and the generic unit for year be y. This gives us:
Let the year unit of measurement for Mercury be ##y_m##
Let the day unit of measurement for Mercury be ##d_m##
Similarly for Earth we have ##y_e ” and ” d_e##
And for Pluto we have ##y_p ” and ” d_p##
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
##color(blue)(“Another potential ambiguity.”)##
Consider the question in parts:
“One year on Mercury is 87.97 on Earth
##color(white)(xxxxxxxxxxx)color(green)(-> 1y_m= 87.97d_e ………………..(1))##
No doubt about the meaning!
“One year on Pluto is 3 times one year on Mercury minus 16.21 days”
Does this mean:##color(white)(…)color(blue)(1y_p=3( 1y_m-16.21d_m)##
Or does it mean:##color(white)(…)color(blue)(1y_p=3( 1y_m)-16.21d_m##
##color(red)(“Assuming this is the case”:color(white)(…)1y_p=3( 1y_m-16.21d_m)…….(2))##
Substitute (1) in (2) giving:
##color(white)(.nnnnnnnnn..)1y_p=3( 87.97d_e-16.21d_m)…….(3)##
~~~~~~~~~~~~~To convert##color(white)(.) 16.21d_m” into ” d_e ##~~~~~~~~~~~~~~~~~~
From (1) ##1y_m=87.97d_e -> 365d_m=87.97d_e##
Divide both sides by 365 giving:
##1d_m=(87.97)/(365) d_e##
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
##color(white)(.nnnnnnnnn..)1y_p=3( 87.97d_e-(16.21 xx87.97)/(365)d_e )…….(3_a)##
I get ##1y_p~~252.19 d_e ## But we need this in years
So ##color(blue)(1y_p~~(252.19)/365y_e ~~ 0.69 y_e)##

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