What is at 6% interest And what is at 14% interest

The rule of the simple interest is
[tex]I=\text{PRT}[/tex]P is the initial amount
R is the rate in decimal
T is the time
Let her put $x in the account of 6%, and $y in the account of 14%
Since the amount in the 14% is more by 900, then
[tex]y=x+900\rightarrow(1)[/tex]Let the first account has I1 interest and the second amount has I2 interest
[tex]I_1=x(R_1)(1)[/tex]Since R1 = 6%
Change it to decimal by dividing it by 100
[tex]R_1=\frac{6}{100}=0.06[/tex]Then
[tex]\begin{gathered} I_1=x(0.06)(1) \\ I_1=0.06x \end{gathered}[/tex]Since R2 = 14%, then
[tex]\begin{gathered} I_2=y(\frac{14}{100})(1) \\ I_2=0.14y \end{gathered}[/tex]Since the total interest is $209, then add I1 and I2 and equate the sum by 209
[tex]0.06x+0.14y=209\rightarrow(2)[/tex]Substitute y in equation (2) by equation (1)
[tex]0.06x+0.14(x+900)=209[/tex]Simplify the left side
[tex]\begin{gathered} 0.06x+0.14(x)+0.14(900)=209 \\ 0.06x+0.14x+126=209 \end{gathered}[/tex]Add the like terms on the left side
[tex]\begin{gathered} (0.06x+0.14x)+126=209 \\ 0.2x+126=209 \end{gathered}[/tex]Subtract 126 from both sides, then divide the 2 sides by 0.2 to find x
[tex]\begin{gathered} 0.2x+126-126=209-126 \\ 0.2x=83 \\ \frac{0.2x}{0.2}=\frac{83}{0.2} \\ x=415 \end{gathered}[/tex]Substitute x in equation (1) by 415
[tex]\begin{gathered} y=900+415 \\ y=1315 \end{gathered}[/tex]6% is $415
14% is $1315