document.write( "Question 869966: Sandra is 36 years old. her son is 14 years old. how many years ago was Sandra 3 times older than her son?\r
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Algebra.Com's Answer #524597 by jim_thompson5910(35256)\"\" \"About 
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Let x = the number of years\r
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\n" ); document.write( "\n" ); document.write( "If we subtract x from each age, we get\r
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\n" ); document.write( "\n" ); document.write( "Sandra's age x yrs ago: 36 - x
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\n" ); document.write( "\n" ); document.write( "We want to know when she was 3 times older, so \r
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\n" ); document.write( "\n" ); document.write( "(Sandra's age x years ago) = 3*(Son's age x yrs ago)\r
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\n" ); document.write( "\n" ); document.write( "36 - x = 3*(14 - x)\r
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\n" ); document.write( "\n" ); document.write( "Now we solve for x\r
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\n" ); document.write( "\n" ); document.write( "36 - x = 3*(14 - x)\r
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\n" ); document.write( "\n" ); document.write( "36 - x = 3*(14) + 3*(-x)\r
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\n" ); document.write( "\n" ); document.write( "36 - x = 42 - 3x\r
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\n" ); document.write( "\n" ); document.write( "36 = 42 - 3x + x\r
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\n" ); document.write( "\n" ); document.write( "36 - 42 = -3x + x\r
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\n" ); document.write( "\n" ); document.write( "-6 = -2x\r
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\n" ); document.write( "\n" ); document.write( "-6/(-2) = x\r
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\n" ); document.write( "\n" ); document.write( "3 = x\r
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\n" ); document.write( "\n" ); document.write( "x = 3\r
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\n" ); document.write( "\n" ); document.write( "So 3 years ago, Sandra was 3 times older than her son (sandra = 33 years, son = 11 years)
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