document.write( "Question 1007101: Three times a first number decreased by a second number is 3.the first number increases by twice the second number is 15.find the numbers \n" ); document.write( "
Algebra.Com's Answer #623094 by addingup(3677)\"\" \"About 
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3x-y= 3
\n" ); document.write( "x+2y= 15 Let's multiply both sides by 3:
\n" ); document.write( "3x+6y= 45 Now subtract this equation from the first:
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\n" ); document.write( "3x-y= 3
\n" ); document.write( "- 3x-6y= -45 (watch your signs, you are subtracting so numbers turn negative)
\n" ); document.write( "___ -7y= -42
\n" ); document.write( "____ y= 6
\n" ); document.write( "So we have that y, the second number, is 6. Now back to the top, to the first equation:
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\n" ); document.write( "3x-y= 3 Using y= 6:
\n" ); document.write( "3x-6= 3
\n" ); document.write( "3x= 9
\n" ); document.write( "x= 3 This is the first number
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\n" ); document.write( "Check:
\n" ); document.write( "3(3)-6= 3
\n" ); document.write( "9-6= 3 OK, we got this correct
\n" ); document.write( "3+2(6)= 15
\n" ); document.write( "3+12= 15
\n" ); document.write( "15=15 This one is also correct. We have the right answers. \n" ); document.write( "
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