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September 19, 2014

Posted by **Mathshelpme** on Monday, January 2, 2012 at 6:50am.

- Maths -
**Reiny**, Monday, January 2, 2012 at 10:25amFrom the first sentence I can tell that J must have $12 more than A

Also from the second sentence A must have at least $6

so form two columns of possible amounts, starting with A having 6

A J

6 18

7 19

8 20

9 21

10 22

11 23

12 24

13 25

14 26

15 27

16 28

17 29

18 30

19 31

...

Now metally take 6 away from A and add 6 to J

when will the first value be 1/3 of the 2nd?

how about 18 and 30

(18-6) = (1/3)(30+6) ----> 12 = (1/3) of 36

I bet you can't wait to learn Algebra. You will be so surprised how really easy these questions become.

J-6 = A+6 ---> J-A = 12

A-6 = (1/3)(J+6)

3A - 18 = J+6

3A - J = 24

3A - J = 24

-A + J = 12

add them

2A = 36

A = 18, then J = A+12 = 30

- Maths -
**Steve**, Monday, January 2, 2012 at 10:48amok. no algebra*

We know that J has $12 more than A. That's because when J is $6 less, it is the same as the new A, which is $6 more.

J can't have just $12, because that would mean that A started with 0, and could not give 6 to J.

A and J must both be multiples of $6, since we're juggling $6 chunks.

So, suppose A = 12. That would make J=24, but 12-6=6 is not 1/3 of 24+6=30.

If A=18, J=30 and 18-6=12 is 1/3 of 30+6=36.

*

But all of this reasoning is just doing algebra in your head. It's never to soon to learn a little algebra, so you might as well start now.

Algebra just gives you a way of writing down number facts so you don't have to keep track of everything in your head.

Let J and A represent the two starting amounts.

J gives A $6, so now we have

J-6 = A+6

If you have a true sentence (or equation), it stays true as long as, if you make changes, you make the same changes to both sides of the equation.

For example, if J is 5 years older than A, she will still be 5 years older a year from now. That is, if

J = A+5

next year,

J+1 = A+5+1

In our problem, we have

J-6 = A+6

If you add 6 to both sides, you get J by herself:

J-6+6 = A+6+6

J = A+12

Now, if we have two unknown numbers, we need to know two independent things about how they fit together. Our second thing is: If A gives J $6, she has 1/3 of J's new amount. Or, avoiding fractions, J will then have 3 times as much as A. In symbols,

J+6 = 3 * (A-6)

Here * means times. To get rid of the parentheses, you need to know about how multiplication distributes over addition. You already know this, but maybe not by name. If someone asks you to add 300 and 400, you have no trouble coming up with 700 as the answer. In your head, you probably say, well considering hundreds, 3+4=7. In symbols,

300 + 400 = 3*100 + 4*100 = (3+4)*100

In other words, you do what's in the parentheses first, then do the multiplication. In our case, we have

J+6 = 3*(A-6)

J+6 = 3A - 3*6

J+6 = 3A - 18

Add 18 to both sides:

J+24 = 3A

Now, we know J=A+12. Wherever we have J, we can just as well use A+12. We can make that substitution, and we now have

A+12 + 24 = 3A

A + 36 = 3A

Now, we need to get all the A's together, so subtract A from both sides.

A-A + 36 = 3A - A

But, A-A = 0, and 3A-A = 2A

So, now we have

36 = 2A

In words, 36 is twice as much as A.

so, A is just half of 36 = 18.

Since J=A+12, J=18+12 = 30.

Congratulations. While there are still some gaps to fill in, you've just covered half a semester of algebra.

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