Question 1184088
<br>
For SOME students, working the problem backwards might be easier....<br>
But solving it "forwards" is a good exercise in algebra, which some students might find more to their liking that solving it backwards.<br><pre>
   Juliet                             Ken
 ----------------------------------------------------------------------
     x                                 y
   (4/5)x                            y+(1/5)x             Juliet gave 1/5 of hers to Ken
  (4/5)x+(2/3)(y+(1/5)x)+4           (1/3)(y+(1/5)x))-4   Ken gave 2/3 of his marbles plus 4 more to Juliet
  (4/5)x+(2/3)(y+(1/5)x)+4-29        (1/3)(y+(1/5)x))-4   Juliet lost 29 marbles</pre>
Juliet ended with 97 marbles:<br>
(4/5)x+(2/3)(y+(1/5)x)+4-29=97
(4/5)x+(2/3)y+(2/15)x=122
(14/15)x+(2/3)y=122
14x+10y=1830 [1]<br>
Ken ended with 21 marbles:<r>
(1/3)(y+(1/5)x))-4=21
(1/3)y+(1/15)x=25
5y+x=375 [2]<br>
Solve [1] and [2] by elimination, dividing [1] by 2:
5y+7x=915
5y+x=375
6x=540
x=90
5y+90=375
5y=285
y=57<br>
ANSWERS:
Juliet started with x=90 marbles
Ken started with y=57 marbles<br>
CHECK:
Start: (J,K)=(90,57)
Juliet gives 1/5 of hers (18) to Ken: (J,K)=(72,75)
Ken gives 2/3 of his plus 4 more (54) to Juliet: (J,K)=(126,21)
Juliet loses 29: (97,21)<br>