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### Best Answer paxwill says

This one wasn't too bad to set up, but I had to use a polynomial solver tool to get the roots of the quartic equation that came out of it. First I called the currect age of the twin girls x and the current boy's age y (like chromosomes, lol). Then I got the two equations

[(y+5)(x+5)^2]/(yx^2) = 6

[(y+10)(x+10)^2]/(yx^2) = 18

Solving each one for y gives you the new equations

y = (5x^2+50x+125)/(5x^2-10x-25)

y = (10x^2+200x+1000)/(17x^2-20x-100)

Setting the right hand sides equal to each other and simplifying gives you a 4th degree polynomial in x:

7x^4 - 30x^3 - 425x^2 + 1500x + 2500 = 0

This quartic equation has 4 real solutions, only one of which is an integer as required by the conditions of the problem. The integer solution is x = 5, which tells you that y = 10. You can double check that their current ages are 5, 5, and 10 by looking at the produts 5 years from now and 10 years from now:

5*5*10 = 250 (current)

10*10*15 = 1500 (6 times 250)

15*15*20 = 4500 (18 times 250)

So the final answer is that the boy is older than his twin sisters. (Almost forget that part!)

Thanks and good explanation.

### DreamerMeg says

I think think requires solving simultaneous equations? I also think I can write them but not sure my elderly brain can solve them!

Daughters = x

Son = y

Z=x^2y

6z = (x+5)^2 (y+5)

18z = (x+10)^2 (y+10)

That's as far as I can get at present. Was going to put it in an Excel spreadsheet and just change the numbers but thought that might be considered cheating!

Meg, you can solve it any way you like, and it's definitely not cheating to use a solving tool. The hardest part is probably setting up the equations. :)

### I'M BANNED Y'ALL!!!!!!!!!! says

Depends on whether or not they are evil twins. Evil twins transcend age.

Good point, that changes everything.

### TinaAtHome says

Are you sure the question is correct? Because if you have 3 numbers and multiply them together, if those numbers are each 5 bigger the new product will be much bigger than 6 times the original number.

Even if their ages are 1, 1 and 1 in 5 years time they will be 6, 6 and 6, the product is 216 and the product will always be bigger than 6 times.

This problem is correctly stated. The twin daughters are not the same age as the son, though they all share a common birthday as noted in the original statement of the problem. :)

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