document.write( "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. \n" ); document.write( "
Algebra.Com's Answer #629130 by macston(5194)\"\" \"About 
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\n" ); document.write( "A=Amy's age; B=Betty's age
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\n" ); document.write( "\"B=2A-5\" Use this to substitute for B below.
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\n" ); document.write( "\"B%5E2-2A%5E2=7\" Substitute (2A-5) for B
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\n" ); document.write( "\"%282A-5%29%5E2-2A%5E2=7\"
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\n" ); document.write( "\"4A%5E2-20A%2B25-2A%5E2=7\"
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\n" ); document.write( "\"2A%5E2-20A%2B18=0\"
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\n" ); document.write( "\"A%5E2-10A%2B9=0\"
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\n" ); document.write( "\"%28A-9%29%28A-1%29=0\"
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\n" ); document.write( "\"A-9=0\" \"OR\" \"A-1=0\"
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\n" ); document.write( "\"A=9\" \"OR\" \"A=1\"
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\n" ); document.write( "Since twice Amy's age is 5 more than Betty's age, 1 does not work:
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\n" ); document.write( "2A-5=B
\n" ); document.write( "2(1)-5=B
\n" ); document.write( "2-5=B
\n" ); document.write( "-3=B
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\n" ); document.write( "And Betty cannot be -3 years old.
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\n" ); document.write( "So Amy is 9 years old.
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\n" ); document.write( "B=2A-5
\n" ); document.write( "B=2(9)-5
\n" ); document.write( "B=16-5=13
\n" ); document.write( "Betty is 13 years old.
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\n" ); document.write( "A+B=9+13=22
\n" ); document.write( "ANSWER: The sum of their ages is 22 years
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\n" ); document.write( "CHECK:
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\n" ); document.write( "\"B%5E2-2A%5E2=7\"
\n" ); document.write( "\"13%5E2-2%289%5E2%29=7%0D%0A%7B%7B%7B169-162=7\"
\n" ); document.write( "\"7=7\"
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