SOLUTION: The sum of the ages of two sisters, Mary and Jane, is 24 years. Four years ago three times Mary’s age was 3 years more than twice Jane’s age. Determine their current ages

Algebra ->  Equations -> SOLUTION: The sum of the ages of two sisters, Mary and Jane, is 24 years. Four years ago three times Mary’s age was 3 years more than twice Jane’s age. Determine their current ages      Log On


   

Question 1170090: The sum of the ages of two sisters, Mary and Jane, is 24 years. Four years ago three
times Mary’s age was 3 years more than twice Jane’s age. Determine their current ages
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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The sum of the ages of two sisters, Mary and Jane, is 24 years.
m + j = 24
j = (24 - m), use this form for substitution
Four years ago three times Mary’s age was 3 years more than twice Jane’s age.
3(m - 4) = 2(j - 4) + 3
3m - 12 = 2j - 8 + 3
3m = 2j - 5 + 12
3m = 2j + 7
replace j with (24-m)
3m = 2(24-m) + 7
3m = 48 - 2m + 7
3m + 2m = 48 + 7
5m = 55
m = 55/5
m = 11 yrs is Mary's age
then
j = 24 - 11
j = 13 yrs is Jane's age
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Check solution in the statement:
"Four years ago three times Mary’s age was 3 years more than twice Jane’s age."
3(11-4) = 2(13-4) + 3
3(7) = 2(9) + 3