Question

Ann is older than Betty. Their ages in years are such that twice the square of Betty's age subtracted from the square of Ann's age gives a number equal to 6 times the difference of their ages. Given also that the sum of their ages is equal to 5 times the difference of their ages, find the age in years if each of the sisters.

This question is actually from my exercise book. The answer shows:

Let Ann be A and Betty be B
A^2-2B^2= 6(A-B)
A+B=5(A-B)=
5A-5B

I don't understand the part where they wrote (a+b)=5(a-b), could someone explain it to me. Much appreciated!

Answer 1

It is just a matter of translating the English to Math.
Going with your definitions of A and B

"the sum of their ages" ---> A + B
"the difference of their ages" --> A-B
"is equal to 5 times" ---> = 5(....

(the sum of their ages) <b> is equal to 5 times</b> the difference of their ages
--> A+B = 5(A-B)
or
A+B = 5A - 5B
6B = 4A
B = (2/3)A or 2A/3

Now do the same analysis for the first part of the question.


so back in A^2 - 2B^2 = 6(A-B) = 6A - 6B
A^2 - 2(2A/3)^2 = 6A - 6(2A/3)
A^2 - 8A^2 /9 = 6A - 4A = 2a
times 9 to clear fractions
9A^2 - 8A^2 = 18A
A^2 - 18A = 0
A(A-18) = 0
A = 0 or A = 18 , A=0 clearly does not work

So Ann is 18 and Betty is 12

check:
square of Ann's age = 324
Twice the square of Betty's age = 288
that difference is 36
6 times the difference of their ages = 6(18-12) = 36
both results are 36, so my answer is correct

Answer 2

Thanks