Question 935064
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Let Dad's age be represented by *[tex \Large d], the boy's age be represented by *[tex \Large b], Mom's age by *[tex \Large m] and each of the twin girls' ages by *[tex \Large g]


Here are some things that we know:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1:\ d\ +\ b\ +\ m\ +\ 2g\ =\ 123]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2:\ d\ +\ b\ =\ 59]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 3:\ m\ +\ 2g\ =\ 59\ +\ 5\ =\ 64]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4:\ d\ =\ m\ +\ 3]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 5:\ b\ =\ g\ +\ 3]


Which is to say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 6:\ g\ =\ b\ -\ 3]


Substituting 6 into 3 we get *[tex \Large m\ +\ 2b\ =\ 70]


Substituting 4 into 2 we get *[tex \Large m\ +\ b\ =\ 56]


The last two equations form a 2X2 system in *[tex \Large m] and *[tex \Large b].  Solve the system by any convenient method to determine the Mom's and the son's ages.  The other two ages, Dad and each of the twin girls are simply a matter of arithmetic from that point.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \