document.write( "Question 1080259: How should a $45,000 investment be split so that the total annual earnings are $1,980, if one portion is invested at 3 percent annual interest amd the rest at 6.5 percent annual interest? \n" ); document.write( "

Algebra.Com's Answer #694440 by josmiceli(19441) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"+I+=+P%2Ar%2At+\"$ ( simple interest )
\n" ); document.write( "Let $\"+a+\"$ = amount invested @ 3%
\n" ); document.write( " $\"+45000+-+a+\"$ = amount invested @ 6.5%
\n" ); document.write( "Let $\"+b+\"$ = earnings from the 3% investment
\n" ); document.write( " $\"+1980+-+b+\"$ = earnings from the 6.5% investment
\n" ); document.write( " $\"+t+=+1+\"$ yr
\n" ); document.write( "------------------------------------------------
\n" ); document.write( "(1) $\"+b+=+a%2A.03%2A1+\"$
\n" ); document.write( "(2) $\"+1980+-+b+=+%28+45000+-+a+%29%2A.065%2A1+\"$
\n" ); document.write( "-------------------------------------------
\n" ); document.write( "(1) $\"+b+=+.03a+\"$
\n" ); document.write( "and
\n" ); document.write( "(2) $\"+1980+-+b+=+2925+-+.065a+\"$
\n" ); document.write( "(2) $\"+b+=+.065a+-+945+\"$
\n" ); document.write( "Plug (1) into (2)
\n" ); document.write( "(2) $\"+.03a+=+.065a+-+945+\"$
\n" ); document.write( "(2) $\"+.035a+=+945+\"$
\n" ); document.write( "(2) $\"+a+=+27000+\"$
\n" ); document.write( "and
\n" ); document.write( " $\"+45000+-+a+=+45000+-+27000+\"$
\n" ); document.write( " $\"+45000+-+a+=+18000+\"$
\n" ); document.write( "---------------------------------
\n" ); document.write( "$27,000 @ 3%
\n" ); document.write( "$18,000 @ 6.5%
\n" ); document.write( "------------------
\n" ); document.write( "check:
\n" ); document.write( "(1) $\"+b+=+a%2A.03%2A1+\"$
\n" ); document.write( "(1) $\"+b+=+27000%2A.03%2A1+\"$
\n" ); document.write( "(1) $\"+b+=+810+\"$
\n" ); document.write( "and
\n" ); document.write( "(2) $\"+1980+-+b+=+%28+45000+-+a+%29%2A.065%2A1+\"$
\n" ); document.write( "(2) $\"+1980+-+b+=+18000%2A.065%2A1+\"$
\n" ); document.write( "(2) $\"+1980+-+b+=+1170+\"$
\n" ); document.write( "(2) $\"+b+=+810+\"$
\n" ); document.write( "OK\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

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