document.write( "Question 170451: Solve the problem. Roberto invested some money at 7%, and then invested $2,000 more than twice this amount at 11%. His total annual income from the two investments was $3,990. How much was invested at 11%? \n" ); document.write( "

Algebra.Com's Answer #125903 by jojo14344(1513) $\"\"$ $\"About$

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Let $\"x\"$ ---> money inv. 7%
\n" ); document.write( "Therefore, $\"2x%2B2000\"$ ---> money inv. @ 11%
\n" ); document.write( "It follows,
\n" ); document.write( " $\"0.07x%2B%282x%2B2000%290.11=3990\"$
\n" ); document.write( " $\"0.07x%2B0.22x%2B220=3990\"$
\n" ); document.write( " $\"0.29x=3990-220=3770\"$
\n" ); document.write( " $\"cross%280.290%29x%2Fcross%280.29%29=cross%283770%2913000%2Fcross%280.29%29\"$
\n" ); document.write( " $\"x\"$ =$13,000, amount inv @ 7%
\n" ); document.write( "Then,
\n" ); document.write( " $\"2%2A13000%2B2000=26000%2B2000\"$ =$28,000, amount inv at 11% (ANSWER)
\n" ); document.write( "Check,
\n" ); document.write( " $\"0.07%2A13000%2B28000%280.11%29=3990\"$
\n" ); document.write( " $\"910%2B3080=3990\"$
\n" ); document.write( " $\"3990=3990\"$
\n" ); document.write( "Thank you,
\n" ); document.write( "Jojo \n" ); document.write( "

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