document.write( "Question 30531: Remember Little Johnny? He now has two investments that produce a $150 income each month. If $1,000 more is invested at 9% than at 10%, how much was invested at each percent?
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Algebra.Com's Answer #17197 by mathtech5(4) $\"\"$ $\"About$

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Let x represent the amount invested in the 10%. Then (x+100) would be the amount invested at 9%. \r
\n" ); document.write( "\n" ); document.write( "Set up the equation: .1x + .09(x+1000)= 150\r
\n" ); document.write( "\n" ); document.write( " .1x + .09x + 100 = 150
\n" ); document.write( " .19x + 100 = 150 (subtract 100 from both sides)
\n" ); document.write( " .19x = 50 (divide by .19)
\n" ); document.write( " x = 263.16\r
\n" ); document.write( "\n" ); document.write( "Therefore, the amount invested at 10% was $263.16 and the amount invested in 9% was $1263.16.
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