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Thelma invested $50,000 in 3 different investments at 6%, 7%, 9%. She invested $20,000 more in 9% than in 6%. If at the end of one year she received interest of $3,970 on her investments, how much did she invest in each of her 3 investments? Thank you for help.

Guest Apr 8, 2018
 #1
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Thelma invested $50,000 in 3 different investments at 6%, 7%, 9%. She invested $20,000 more in 9% than in 6%. If at the end of one year she received interest of $3,970 on her investments, how much did she invest in each of her 3 investments? Thank you for help.

 

http://free-picload.com/index.php?seite=upload.5e7zn

http://free-picload.com/index.php?seite=upload.5e7zn

 

laugh

Omi67  Apr 8, 2018
edited by Omi67  Apr 8, 2018
edited by Omi67  Apr 8, 2018
 #2
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Omi67: What is the link you gave for? It is for uploading pictures...etc. This question is about algebraic equations and how to solve them. At any rate, thank you.

Guest Apr 8, 2018
 #3
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Let her investment in 6% =S
Let her investment in 7% =T
Her investment in 9% (N) =S + 20,000 
S + T + S + 20,000 = 50,000.................................................................(1)
0.06S + 0.09(S+20,000) +  .07*T  = 3970, solve for S, T......................(2) 
From(1) above, T =50,000 - 20,000 - 2S=30,000 - 2S. Sub this for T in(2) and you should get:

0.06S + .09S + 1,800 + 0.07(30,000 - 2S) =3,970

0.15S + 1,800 + 2,100 - 0.14S = 3,970

0.01S + 3,900 = 3,970

0.01S = 3,970 - 3,900

0.01S = 70. Therefore, Thelma's investments are:
S=$7,000 @ 6%,  T=$16,000 @ 7%,   N=$7000+$20000 =$27,000 @ 9%.  

Guest Apr 8, 2018
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Click on the blue link and then click on the image on the top left

Omi67  Apr 8, 2018

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