SOLUTION: The sum of 3 times Don's age and 5 times Sean's age is 150. Sean is 9 years less than twice as old as Don is. What are each of their ages?

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Question 610998: The sum of 3 times Don's age and 5 times Sean's age is 150. Sean is 9 years less than twice as old as Don is. What are each of their ages?
Answer by Maths68(1474) About Me  (Show Source):
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Let
Present age of Don = d
Present age of Sean = s

The sum of 3 times Don's age and 5 times Sean's age is 150
3d+5s=150...............(1)

Sean is 9 years less than twice as old as Don is
s=2d-9..................(2)

Put the value of s from (2) to (1)
3d+5s=150...............(1)
3d+5(2d-9)=150
3d+10d-45=150
13d=150+45
13d=195
13d/13=195/13
d=15
Put the value of d in (2)
s=2d-9
s=2(15)-9
s=30-9
s=21

Present age of Don = d = 15
Present age of Sean = s = 21


Check
========
The sum of 3 times Don's age and 5 times Sean's age is 150
3d+5s=150
3(15)+5(21)=150
45+105=150
150=150
Sean is 9 years less than twice as old as Don is
s=2d-9
21=2(15)-9
21=30-9
21=21