document.write( "Question 1015721: Larry Mitchell invested part of his $21,000 advance at 8% annual simple interest and the rest at 7% annual simple interest. If his total yearly interest from both accounts was $1,580, find the amount invested at each rate. \n" ); document.write( "

Algebra.Com's Answer #632116 by addingup(3677) $\"\"$ $\"About$

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Some at 0.08
\n" ); document.write( "Some at 0.07
\n" ); document.write( "Total 21,000
\n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\n" ); document.write( "Let's call the investment at 8% x
\n" ); document.write( "What's left of the 21,000 is at 7%
\n" ); document.write( "So:
\n" ); document.write( "0.08 = x
\n" ); document.write( "0.07 = 21,000-x
\n" ); document.write( "0.08x+0.07(21,000-x) = 1,580
\n" ); document.write( "0.08x+1,470-0.07x = 1,580 Subtract x on left and subtract 1,470 on both sides:
\n" ); document.write( "0.01x = 110 Divide both sides by 0.01
\n" ); document.write( "x = 11,000 is at 8%. And:
\n" ); document.write( "21,000-11,000 = 10,000 at 7%
\n" ); document.write( "~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
\n" ); document.write( "Check:
\n" ); document.write( "0.08*11,000 = 880
\n" ); document.write( "0.07*10,000 = 700
\n" ); document.write( "___________-------
\n" ); document.write( "Total... = 1,580 We have the correct answer \n" ); document.write( "

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