document.write( "Question 1103689: Mr. Traynor invested a sum of money at 6%. He invested a second sum, $250 more than the first sum, at 8%. If his total annual income was $300, how much did he invest? \n" ); document.write( "
Algebra.Com's Answer #718401 by richwmiller(17219)\"\" \"About 
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We don't know the total invested.
\n" ); document.write( "We know the difference in the two accounts.
\n" ); document.write( "We know that the interest for the two accounts is $300
\n" ); document.write( "0.08*x+0.06*y=300
\n" ); document.write( "We know that the account at 8% has $250 more.
\n" ); document.write( "x=250+y
\n" ); document.write( "We substitute for x
\n" ); document.write( "0.08*(250+y)+0.06*y=300
\n" ); document.write( "We multiply out
\n" ); document.write( "20+0.08y+0.06*y=300
\n" ); document.write( "We combine like terms.
\n" ); document.write( "0.14*y=280
\n" ); document.write( "Isolate y
\n" ); document.write( "y=$2000 at 6%
\n" ); document.write( "x=250+y
\n" ); document.write( "Calculate x
\n" ); document.write( "x=$2250 at 8%
\n" ); document.write( "Total invested $2250+$2000=$4250
\n" ); document.write( "Now,we know the total invested is: $4250
\n" ); document.write( "We check
\n" ); document.write( "0.08*2250+0.06*2000=300
\n" ); document.write( "180+120=300
\n" ); document.write( "300=300
\n" ); document.write( "Since this statement is TRUE and neither x nor y is negative all is well.
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