Question 193356
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Let <b><i>x</i></b> represent the age of the one doing the talking.  If you take  two years off of that age you have <b><i>x</i> - 2</b>.  Let <b><i>y</i></b> represent the age of the the other one.  If you add two years to that age you have <b><i>y</i> + 2</b>.


After the one doing the talking has given away the two years:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(x - 2) = y + 2]


Likewise, if you take three years off of the talker's age you have <b><i>x</i> - 3</b>.   If you add two years to the other one's age you have <b><i>y</i> + 3</b>.  And:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3(x - 3) = y + 3]


Put the equations in standard form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x - y = 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x - y = 12]


Multiply the first equation by -1 and then add the equations:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -2x + y = -6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3x - y = 12]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x + 0y = 6]


The one doing the talking is 6, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2(6) - y = 6]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = 6]


And the other one is 6 as well.  Hmmm, twins, and mighty clever for 6 year olds.


Check:


6 minus 2 is 4, 6 plus 2 is 8, 8 is twice 4.


6 minus 3 is 3, 6 plus 3 is 9, 9 is three times 3.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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