document.write( "Question 1123204: A committee is to have 3 more women than men and the number on the committee must be at least 7 but no more than 15. What are the possible numbers of women on the committee? \n" ); document.write( "
Algebra.Com's Answer #739563 by greenestamps(13203)\"\" \"About 
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\n" ); document.write( "(1) Algebraically....

\n" ); document.write( "We are asked to find the possible numbers of women on the committee, so use that as our variable; the number of men will be 3 less:

\n" ); document.write( "let x = number of women
\n" ); document.write( "then x-3 = number of men

\n" ); document.write( "The total number of people on the committee -- x, plus (x-3) -- is to be between 7 and 15:

\n" ); document.write( "7 <= x+(x-3) <= 15
\n" ); document.write( "7 <= 2x-3 <= 15
\n" ); document.write( "10 <= 2x <= 18
\n" ); document.write( "5 <= x <= 9

\n" ); document.write( "The number of women can be between 5 and 9.

\n" ); document.write( "Using logical reasoning....

\n" ); document.write( "Not counting the 3 \"extra\" women, the committee consists of equal numbers of men and women, and the total number on the committee is between 4 and 12; that means the numbers of men and women are between 4/2=2 and 12/2=6 each. And then counting the other 3 women, the number of women is between 2+3=5 and 6+3=9.
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