document.write( "Question 760154: Solve the problem:\r
\n" ); document.write( "\n" ); document.write( "One number is 1 less than a second number. Twice the second number is
\n" ); document.write( "20 less than 4 times the first. Find the two numbers.\r
\n" ); document.write( "\n" ); document.write( "Please show steps on this. Thank you :)
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Algebra.Com's Answer #462468 by MathLover1(20849)\"\" \"About 
You can put this solution on YOUR website!

\n" ); document.write( "if one number (let it be \"x\") is \"1\" less than a second number (let it be \"y\"), then\r
\n" ); document.write( "\n" ); document.write( " \"x%2B1=y\"....eq.1\r
\n" ); document.write( "\n" ); document.write( "if twice the second number is \"20\" less than \"4\" times the first, then\r
\n" ); document.write( "\n" ); document.write( "\"2y%2B20=4x\"....eq.2\r
\n" ); document.write( "\n" ); document.write( "solve the system:\r
\n" ); document.write( "\n" ); document.write( " \"x%2B1=y\"....eq.1\r
\n" ); document.write( "\n" ); document.write( "\"2y%2B20=4x\"....eq.2
\n" ); document.write( "_________________________\r
\n" ); document.write( "\n" ); document.write( " \"x%2B1=y\"....eq.1...substitute in eq.2\r
\n" ); document.write( "\n" ); document.write( "\"2%28x%2B1%29%2B20=4x\"....eq.2\r
\n" ); document.write( "\n" ); document.write( "\"2x%2B2%2B20=4x\"\r
\n" ); document.write( "\n" ); document.write( "\"22=4x-2x\"\r
\n" ); document.write( "\n" ); document.write( "\"22=2x\"\r
\n" ); document.write( "\n" ); document.write( "\"22%2F2=x\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%2811=x%29\"\r
\n" ); document.write( "\n" ); document.write( "now find \"x%2B1=y\"....eq.1\r
\n" ); document.write( "\n" ); document.write( "\"11%2B1=y\"\r
\n" ); document.write( "\n" ); document.write( "\"highlight%2812=y%29\"\r
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