SOLUTION: The sum of 4 times Joan's age and 3 times Jim's age is 47. Jim is 1 year less than twice as old as Joan. Find each of their ages.

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Question 179292: The sum of 4 times Joan's age and 3 times Jim's age is 47. Jim is 1 year less than twice as old as Joan. Find each of their ages.
Found 2 solutions by josmiceli, solver91311:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Let a= Joan's age
Let b= Jim's age
given:
(1) 4a+%2B+3b+=+47
(2) b+=+2a+-+1
substitute b in (2)
for b in (1)
(1) 4a+%2B+3b+=+47
4a+%2B+3%2A%282a+-+1%29+=+47
4a+%2B+6a+-+3+=+47
10a+=+50
a+=+5
And, since
(2) b+=+2a+-+1
b+=+2%2A5+-+1
b+=+9
Joan is 5 and Jim is 9

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


Let Joan's age be represented by x. Then four times Joan's age would be 4x.

Let Jim's age be represented by y. Then three times Jim's age would be 3y.

Four times Joan's age (4x), the sum of (+), three times Jim's age (3y) is (=) 47. So:

Jim is (y =) one year less than twice as old as Joan (2x - 1). So:

Substitute 2x - 1 for y in











Check:

Answer checks.


John