document.write( "Question 381158: Please help me!!!!\r
\n" ); document.write( "\n" ); document.write( "one number is 3 less than a second number. Twice the second number is 3 less than 3 times the first. Find the two numbers.\r
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\n" ); document.write( "gloria
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Algebra.Com's Answer #270427 by Jk22(389)\"\" \"About 
You can put this solution on YOUR website!
Let write number one n1, and the 2nd n2, we translate the problem :
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\n" ); document.write( "n2 - 3 = n1
\n" ); document.write( "2n2 = 3n1 - 3=3(n1-1)
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\n" ); document.write( "if we make equation 1 times 2 minus equation 2, we get rid of the unknown n2 : \r
\n" ); document.write( "\n" ); document.write( "-6 = 2n1 - 3n1 + 3
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\n" ); document.write( "-9 = -n1
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\n" ); document.write( "hence the first number is 9, the second is 12
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\n" ); document.write( "In words the problem would be solved like : twice a number is 6 less than twice a 2nd number, and twice the 2nd number is 3 less three time the number
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\n" ); document.write( "hence twice a number is 6 less than 3 times this number less 3
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\n" ); document.write( "or : twice a number is 3 times a number less 9
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\n" ); document.write( "so twice this number plus 9 is 3 times this number, or finally : this number is 9, the second is 12
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\n" ); document.write( "Verification : 9 = 12 -3 and 2*12=24 and 3*9-3=27-3 = 24
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