SOLUTION: A wife and husband, Linda and Mark, have ages that total 80 years. Two years ago, 2 times Mark's age was 2 years more than 3 times Linda's age. How old is each now?

Question 29327: A wife and husband, Linda and Mark, have ages that total 80 years. Two years ago, 2 times Mark's age was 2 years more than 3 times Linda's age. How old is each now?
Answer by sdmmadam@yahoo.com(530) (Show Source): You can put this solution on YOUR website!
A wife and husband, Linda and Mark, have ages that total 80 years. Two years ago, 2 times Mark's age was 2 years more than 3 times Linda's age. How old is each now?
Let Linda's age be L years
and let Mark's age be M year.
Linda and Mark, have ages that total 80 years.
That is L+M = 80 ---(1)
Two years ago Linda was (L-2) years old
and Mark was (m-2)years old
Then 2 times Mark's age was 5 years more than 3 times Linda's age
That is 2(M-2)= 3(L-2)+2 ----(2)
That is 2M-4=3L-6+2
2M-3L= 4-6+2
2M-3L= 0 ----(*)
using (1) and substituting for L=(80-M)in (*)
2M-3(80-M) =0
2M-240-3M= 0
2M+3M= 240
5M=240
M= 240/5 = 48
L= 80 - 48 = 32
Answer: Mark is now 48 years old and Linda is 32 years old
Verification: Two years ago,M= (48-2) = 46 and L= (32-2) = 30
Twice (46) should be 2 more than thrice(30)
that is 92 should be 2 more than 90 which is so
Hence their ages are correct.
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