document.write( "Question 61422: John Craig invested a portion of $15,000 at 10% annual interest and the balance at 6% annual interest. If he earned $1260 for the year from the two accounts, how much did he invest at 10%? \n" ); document.write( "
Algebra.Com's Answer #42340 by tutorcecilia(2152)\"\" \"About 
You can put this solution on YOUR website!
Ax+By=C [use the standard form of a line]
\n" ); document.write( "10%x+6%y=$1260
\n" ); document.write( ".10x+.06y=1260 [convert the percents to decimals]
\n" ); document.write( "and
\n" ); document.write( "x+y=$15,000
\n" ); document.write( "x+($15,000-x)=$15,000
\n" ); document.write( "putting it all together gives:
\n" ); document.write( ".10x+.06(15,000-x)=1260 [substitute (15,000-x) for y]
\n" ); document.write( ".10x+900-.06=1260
\n" ); document.write( ".04x=360
\n" ); document.write( "x=$9000 invested at 10%
\n" ); document.write( "y=(15,000-x)=15,000-9000=$6,000 invested at 6%
\n" ); document.write( "because $6,000+$9,000=$15,000
\n" ); document.write( "and
\n" ); document.write( ".10(9000)=$900
\n" ); document.write( ".06(6,000)=$360
\n" ); document.write( "900+360=$1260\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
\n" );