SOLUTION: Twice Amy's age is 5 years more than that of Betty's. Twice the square of Amy's age subtracted from the square of Betty's age is 7 years. Find the sum of their ages.

Question 1012958: Twice Amy's age is 5 years more than that of Betty's. Twice the square of Amy's age subtracted from the square of Betty's age is 7 years. Find the sum of their ages.
Answer by macston(5194) (Show Source): You can put this solution on YOUR website!
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A=Amy's age; B=Betty's age
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Use this to substitute for B below.
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Substitute (2A-5) for B
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Since twice Amy's age is 5 more than Betty's age, 1 does not work:
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2A-5=B
2(1)-5=B
2-5=B
-3=B
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And Betty cannot be -3 years old.
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So Amy is 9 years old.
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B=2A-5
B=2(9)-5
B=16-5=13
Betty is 13 years old.
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A+B=9+13=22
ANSWER: The sum of their ages is 22 years
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CHECK:
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