document.write( "Question 1099311: A father wishes to share $9000 between his two sons so that a fifth part of the eldest son's share exceeds by 90, the fourth part of what goes to the younger son. Calculate what each son would receive.
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Algebra.Com's Answer #713751 by Theo(13342)\"\" \"About 
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x equals the older son's share.
\n" ); document.write( "y equals the younger son's share.\r
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\n" ); document.write( "\n" ); document.write( "x + y = 9000\r
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\n" ); document.write( "\n" ); document.write( "1/5 * x = 1/4 * y + 90\r
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\n" ); document.write( "\n" ); document.write( "multiply both side of the second equation by 5 to get:\r
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\n" ); document.write( "\n" ); document.write( "x = 5/4 * y + 450\r
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\n" ); document.write( "\n" ); document.write( "replace x in the first equation with this to get:\r
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\n" ); document.write( "\n" ); document.write( "5/4 * y + 450 + y = 9000\r
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\n" ); document.write( "\n" ); document.write( "combine like terms and subtract 450 from both sides of the equation to get:\r
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\n" ); document.write( "\n" ); document.write( "9/4 * y = 8550\r
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\n" ); document.write( "\n" ); document.write( "multiply both sides of this equation by 4/9 to get:\r
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\n" ); document.write( "\n" ); document.write( "y = 3800\r
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\n" ); document.write( "\n" ); document.write( "since x + y = 9000, then x = 5200\r
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\n" ); document.write( "\n" ); document.write( "you have x = 5200 and y = 3800 for a total of 9000\r
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\n" ); document.write( "\n" ); document.write( "1/5 * x = 1/4 * y + 90 becomes 1/5 * 5200 = 1/4 * 3800 + 90 which becomes 1040 = 950 + 90 which becomes 1040 = 1040 which is true.\r
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\n" ); document.write( "\n" ); document.write( "your solution is that the older son receives 5200 and the younger son receives 3800.\r
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