SOLUTION: The ages in years of three sisters are consecutive multiples of five. Six years ago the sum of their ages was 72. Find their present ages.
I have gotten this far:
Let x= the
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Question 540688: The ages in years of three sisters are consecutive multiples of five. Six years ago the sum of their ages was 72. Find their present ages.
I have gotten this far:
Let x= the youngest sisters age now
x+5 = the middle sisters age now
xt10 = the oldest sisters age now
(x-6)+(x-1)+(x+4)=72
3x+3 =72
-3+3x+3 =72+-3
0+3x =69
3x =69
1/3×3x =69×1/3
x =23
Is this correct?
Thank you for helping me.
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
Let x= the youngest sisters age now
x+5 = the middle sisters age now
x+10 = the oldest sisters age now
(x-6)+(x-1)+(x+4)=72
3x-3 =72
-3+3x+3 =72+3
0+3x =75
3x =75
1/3×3x =75×1/3
x =25
25, 30,35
CHECK
6 years ago
19,24,29
add up
72
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