document.write( "Question 16966: Different interest rates. Mrs. Brighton invested $30,000 and received a total of #2,300 in interest. If she invested part of the money at 10% and the remainder at 5%, then how much did she invest at each rate? \n" ); document.write( "

Algebra.Com's Answer #8246 by Earlsdon(6294) $\"\"$ $\"About$

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Let x = the amount invested at 10% and ($30,000 - x) the amount invested at 5%
\n" ); document.write( "The interest Mrs. Brighton received from these two investments would be:\r
\n" ); document.write( "\n" ); document.write( "(10% of $x) + (5% of ($30,000 - $x)) and this = $2,300 So, knowing this we can solve for x, the amount invested at 10%, and once we have that, we can find the amount invested at 5%. Let's set up the appropriate equation after changing the percents to their decimal equivalents: 10% = 0.1 and 5% = 0.05\r
\n" ); document.write( "\n" ); document.write( " $\"0.1x+%2B+0.05%2830000+-+x%29+=+2300\"$ Simplify and solve for x.
\n" ); document.write( " $\"0.1x+%2B+1500+-+0.05x+=+2300\"$ Combine the x-terms.
\n" ); document.write( " $\"0.05x+%2B+1500+=+2300\"$ Subtract 1500 from both sides.
\n" ); document.write( " $\"0.05x+=+800\"$ Divide both sides by 0.05
\n" ); document.write( " $\"x+=+16000\"$ \r
\n" ); document.write( "\n" ); document.write( "$16,000 was invested at 10%
\n" ); document.write( "$30,000 - $16,000 = $14,000 was invested at 5%\r
\n" ); document.write( "\n" ); document.write( "Check:
\n" ); document.write( "0.1($16,000) + 0.05($14,000) = $1,600 + $700 = $2,300 Total interest received.
\n" ); document.write( " \n" ); document.write( "

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