document.write( "Question 142840: Dorian had some red and blue beads. he had 24 more red beads than blue beads. he gave away 2/3 of the red beads and 1/4 of the blue beads. Altogether, he gave away 104 beads. how many red beads did he have at first? \n" ); document.write( "
Algebra.Com's Answer #103934 by jojo14344(1513)\"\" \"About 
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Okay, let's assigned \"x\" = # of blue beads. Likewise, knowing he had 24 more red beads than blue beads, so \"x+24\"= red beads, right?
\n" ); document.write( "Another condition, 2/3 of the red beads and 1/4 of blue beads were given away that totals to 104. To show in eqn,
\n" ); document.write( "2/3(red beads) + 1/4(blue beads) = 104 ---------- eqn 1
\n" ); document.write( "2/3(x+24) + 1/4(x) =104 got it? ------------------ eqn 1.1
\n" ); document.write( "continuing,
\n" ); document.write( "2x/3 + 48/3 +1/4(x) =104
\n" ); document.write( "(2/3)(x) + (1/4)(x) =104-16
\n" ); document.write( "(8x+3x)/12 = 88
\n" ); document.write( "11x = 1056
\n" ); document.write( "x=96, Total # of blue beads at first
\n" ); document.write( "Also, x+24 = 96+24 = 120, total # of red beads at first
\n" ); document.write( "In doubt? go back eqn 1,
\n" ); document.write( "2/3(120) + 1/4(96) = 104
\n" ); document.write( "80+24=104
\n" ); document.write( "104=104
\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo
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