SOLUTION: Mr. Jackson is 13 years less than three times as old as his son. Six years ago he was fourteen years more than twice as old as his son was then. Find each of their ages today
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Question 515993: Mr. Jackson is 13 years less than three times as old as his son. Six years ago he was fourteen years more than twice as old as his son was then. Find each of their ages today Answer by Maths68(1474) (Show Source):
Mr. Jackson is 13 years less than three times as old as his son.
j=3s-13........(1)
Six years ago he was fourteen years more than twice as old as his son was then.
j-6=2(s-6)+14
j-6=2s-12+14
j-6=2s+2
j=2s+2+6
j=2s+8..............(2)
Put the value of j from (1) to (2)
3s-13=2s+8
3s-2s=8+13
s=21
Put the value of s in (1)
j=3s-13........(1)
j=3(21)-13
j=63-13
j=50
Present age of Mr. Jackson = j = 50 years old
Present age of Son = s = 21 years old