document.write( "Question 9828: Mary is 20 years old. Her sister, Sara, is 5 years old. In how many years will Mary be twice as old as Sara? \n" ); document.write( "

Algebra.Com's Answer #5307 by prince_abubu(198) $\"\"$ $\"About$

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It's one of those now and later type problems. I guess a good way to begin is to write their ages in a fraction. We'll let Mary's age be in the numerator, and Sara's age in the denominator. So far, we have $\"+20%2F5+\"$ . In 1 year, both of them will be 1 year older. Duh, right? In 2 years, BOTH of them will be two years older. Again, duh. In x years, BOTH of them will be x years older. AHH! How far into the future will we have to go until Mary's age is double Sara's?\r
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\n" ); document.write( "\n" ); document.write( " $\"+%2820%2Bx%29%2F%285%2Bx%29+=+2%2F1+\"$ <--------- Set up a proportion. Since both will be older by the same amount in any time in the future, we add x top both Sarah and Mary's ages. We'll set this equal to 2/1 because we want Mary's age to be double Sarah's\r
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\n" ); document.write( "\n" ); document.write( " $\"+20%2Bx+=+2%285%2Bx%29+\"$ <------- Cross-multiply\r
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\n" ); document.write( "\n" ); document.write( " $\"+20%2Bx+=+10%2B2x+\"$ <------ Distribute\r
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\n" ); document.write( "\n" ); document.write( " $\"+20+=+10+%2B+x+\"$ <-------- subtracted x from both sides.\r
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\n" ); document.write( "\n" ); document.write( " $\"+10+=+x+\"$ <------ subtracted 10 from both sides. It's alright that the variable ends up on the right side. OK. The results say that in 10 years, Mary's age will be double Sara's age.\r
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\n" ); document.write( "\n" ); document.write( "Let's check. In 10 years, Mary will be 30. In 10 years, Sara will be 15. Is 30 double 15? Yes. We got it! \n" ); document.write( "

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