document.write( "Question 1101288: One positive number is 10 less than another positive number. If the reciprocal of the smaller number is added to three times the reciprocal of the larger number, the sum is 1/4. Find the two numbers \n" ); document.write( "
Algebra.Com's Answer #715917 by ankor@dixie-net.com(22740)\"\" \"About 
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One positive number is 10 less than another positive number.
\n" ); document.write( "a = b- 10
\n" ); document.write( " If the reciprocal of the smaller number is added to three times the reciprocal of the larger number, the sum is 1/4.
\n" ); document.write( "\"1%2Fa\" + \"3%2Fb\" = \"1%2F4\"
\n" ); document.write( "multiply by 4ab, cancel the denominators
\n" ); document.write( "4b + 12a = ab
\n" ); document.write( "replace a with (b-10)
\n" ); document.write( "4b + 12(b-10) = b(b-10)
\n" ); document.write( "4b + 12b - 120 = b^2 - 10b
\n" ); document.write( "16b - 120 = b^2 - 10b
\n" ); document.write( "combine to form a quadratic equation on the right
\n" ); document.write( "0 = b^2- 10b - 16b + 120
\n" ); document.write( "b^2 - 26b + 120 = 0
\n" ); document.write( "factors to
\n" ); document.write( "(b-6)(b-20) = 0
\n" ); document.write( "two solutions
\n" ); document.write( "b = 6, not possible, a has to be positive
\n" ); document.write( "b = 20, this is our solution
\n" ); document.write( "then
\n" ); document.write( "a = 20 - 10
\n" ); document.write( "a = 10
\n" ); document.write( " Find the two numbers: 10 & 20
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( ":
\n" ); document.write( "see if that checks out
\n" ); document.write( "\"1%2F10\" + \"3%2F20\" = \"1%2F4\"
\n" ); document.write( "\"2%2F20\" + \"3%2F20\" = \"1%2F4\"
\n" ); document.write( "\"5%2F20\" = \"1%2F4\"
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