SOLUTION: Bill"s age is a year less than twice Jane's age. 5 years ago Bills's age was 3 times Jane's age. Find Bill's age?

Question 1204169: Bill"s age is a year less than twice Jane's age. 5 years ago Bills's age was 3 times Jane's age. Find Bill's age?
Found 3 solutions by Edwin McCravy, greenestamps, ikleyn:
Answer by Edwin McCravy(20054)   (Show Source): You can put this solution on YOUR website!

Word problems are much easier to set up using more than one unknown. But some people think it's easier with only one unknown. I'm sure another tutor will show you how to do it with only one unknown, but it's always harder to set up. Bill"s age is a year less than twice Jane's age. 5 years ago Bills's age was 3 times Jane's age. Substitute 2J-1 for B in the 2nd equation: Substitute 9 for J in Check the words: Bill"s age (17) is a year less than twice Jane's age. Twice Jane's age is 18, and sure enough Bill is a year less. 5 years ago Bills's age (12) was 3 times Jane's age (4), Sure enough, 12 is 3 times 4. Edwin

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!

I very much disagree with the other tutor when he says solving a problem like this is easier with two unknowns than with one.

Compare the algebra required to solve the problem with his solution using two unknowns to that required in the solution below using one unknown.

Using the information that 5 years ago Bill's age was 3 times Jane's age....

Let x = Jane's age 5 years ago
Then 3x = Bill's age 5 years ago

Then their ages now are x+5 and 3x+5.

Bill's age now is 1 less than twice Jane's age now:

Their ages 5 years ago were x=4 and 3x=12.

Their ages now are 4+5=9 and 12+5=17.

ANSWER: Bill's age now is 17

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

Edwin, even although some problems are easier to setup using more than one unknown,

it does not necessary mean that it is easier to SOLVE them using more than one unknown.

So, it may happen, that the setup is easier this way, but the solution is harder.

From this point of view, it is unclear for what position do you fight - and WHY.

There are many problems, that can be stupidly considered as for three unknowns,
but smartly and elegantly solved using only one unknown - I solved many such problems at this forum.

Isn't it is better to develop students' mind to be flexible enough (when possible),
in order for they be able to seek their best approaches to their problems ?

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