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Sagot :
Let x and y be the amounts invested in the one paying 9% and the one paying 10% respectively.
x + y = 14000 ------------------------------------------(1)
[tex]\text{ simple interest = }\frac{\text{ principal}\times\text{ rate}\times\text{ time}}{100}[/tex]For the one paying 9% simple interest
[tex]\text{ simple interest =}\frac{x\times9\times1}{100}=\frac{9x}{100}[/tex]and
for the one paying 10% simple interest
[tex]\text{ simple interest =}\frac{y\times10\times1}{100}=\frac{10y}{100}[/tex]Hence,
[tex]\text{ total interest = }\frac{9x}{100}+\frac{10y}{100}=1339[/tex]Therefore,
[tex]\begin{gathered} \frac{9x+10y}{100}=1339 \\ \Rightarrow9x+10y=133900---------------------(2) \end{gathered}[/tex]From equation (1), isolating x, we have
x = 14000 - y --------------------------------(3)
Substituting equation (3) into equation (2), we have
[tex]\begin{gathered} 9(14000-y)+10y=133900 \\ \text{ Hence} \\ 126000-9y+10y=133900 \\ \Rightarrow y=133900-126000=7900 \\ \end{gathered}[/tex]Substituting the y = 7900 into equation (3), we have
x = 14000 - 7900 = 6100
Hence,
he invested $6100 in the account with 9% simple interest
and $7900 in the account with 10% simple interest
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