document.write( "Question 10124: three consecutive integers are such that the first plus one half the second plus seven less than twice the third is 2101. What are the integers \n" ); document.write( "
Algebra.Com's Answer #5387 by bonster(299)\"\" \"About 
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there are 3 consecutive inteters:
\n" ); document.write( "#1: x
\n" ); document.write( "#2: x+1
\n" ); document.write( "#3: x+2\r
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\n" ); document.write( "\n" ); document.write( "x+\"%281%2F2%29%28x%2B1%29\"+\"%282%29%28x%2B2%29-7\"=2101
\n" ); document.write( "x+\"%28x%2F2%29%2B%281%2F2%29\"+2x+4-7=2101
\n" ); document.write( "2(x+\"x%2F2\"+\"1%2F2\"+2x+4-7=2101)
\n" ); document.write( "2x+x+1+4x+8x-14=4202
\n" ); document.write( "15x-13=4202
\n" ); document.write( "15x=4215
\n" ); document.write( "x=\"4215%2F15\"
\n" ); document.write( "x=281\r
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\n" ); document.write( "\n" ); document.write( "the first number is 281. since they are consecutive numbers, the numbers are 281, 282, 283
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