document.write( "Question 910034: Hello Tutors:here's my question...The total of two numbers is 41...One of the number is 19 less than 3 times the other...Find both numbers. ok this is what I did.. I put x+y=41..then I did the second part as x=3y-19.i got x=26 and y=15
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Algebra.Com's Answer #552211 by MathLover1(20850)\"\" \"About 
You can put this solution on YOUR website!

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\n" ); document.write( "\n" ); document.write( "The total of two numbers is \"41\"...=> \"x%2By=41\".....eq.1\r
\n" ); document.write( "\n" ); document.write( "One of the number, let's say \"x\" is \"19\" less than \"3\" times the other,let's say \"y\" \r
\n" ); document.write( "\n" ); document.write( "\"x\" is \"19\" less than \"3y\" means add \"19\" to \"x\" to make it equal to \"3y\"\r
\n" ); document.write( "\n" ); document.write( " \"x%2B19=3y\" ....eq.2 ..solve for \"x\" and plug it in eq.1\r
\n" ); document.write( "\n" ); document.write( "\"x=3y-19\"\r
\n" ); document.write( "\n" ); document.write( "\"x%2By=41\".....eq.1\r
\n" ); document.write( "\n" ); document.write( "\"3y-19%2By=41\" .....solve for \"y\" \r
\n" ); document.write( "\n" ); document.write( "\"4y-19=41\"\r
\n" ); document.write( "\n" ); document.write( "\"4y=41%2B19\"\r
\n" ); document.write( "\n" ); document.write( "\"4y=60\"\r
\n" ); document.write( "\n" ); document.write( "\"y=60%2F4\"\r
\n" ); document.write( "\n" ); document.write( "\"y=15\"\r
\n" ); document.write( "\n" ); document.write( "now find \"x\"\r
\n" ); document.write( "\n" ); document.write( "\"x=3y-19\"\r
\n" ); document.write( "\n" ); document.write( "\"x=3%2A15-19\"\r
\n" ); document.write( "\n" ); document.write( "\"x=45-19\"\r
\n" ); document.write( "\n" ); document.write( "\"x=26\"\r
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