document.write( "Question 1162082: Mr. Wilson invested money in two accounts. His total investment was $29,000. If one account pays 4% in interest and the other pays 7% in interest, how much did he invest in each account if he earned a total of $1,550 in interest in 1 year? \n" ); document.write( "
Algebra.Com's Answer #785794 by greenestamps(13200)\"\" \"About 
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\n" ); document.write( "A setup using the traditional algebraic approach would look something like this:

\n" ); document.write( "let x = amount invested at 4%
\n" ); document.write( "then 29000-x = amount invested at 7%

\n" ); document.write( "The interest from one investment is .04(x); from the other is .07(29000-x).

\n" ); document.write( "The total interest was $1160:

\n" ); document.write( "\".04%28x%29%2B.07%2829000-x%29+=+1160\"

\n" ); document.write( "Solve using basic algebra. I leave that to you.

\n" ); document.write( "I find this alternative method faster and easier....

\n" ); document.write( "(1) Determine how the actual interest compares to the interest that would have been earned if the whole $29000 had been invested at each rate.
\n" ); document.write( "all at 4%: .04(29000) = 1160
\n" ); document.write( "actual interest: 1550
\n" ); document.write( "all at 7%: .07(29000) = 2030

\n" ); document.write( "(2) Where the actual interest lies between the two extremes exactly determines what fraction of the total was invested at each rate.

\n" ); document.write( "The difference between 1160 and 2030 is 870
\n" ); document.write( "The difference between 1160 and 1550 is 390
\n" ); document.write( "The actual interest 1550 is 390/870 = 39/87 = 13/29 of the way from 1160 to 2030.

\n" ); document.write( "That means 13/29 of the total was invested at the higher rate.

\n" ); document.write( "ANSWER: 13/29 of $29,000, or $13,000, at 7%; the other $16,000 at 4%.

\n" ); document.write( "CHECK: .07(13000)+.04(16000) = 910+640 = 1550

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