Connor had $91.00 to invest. Part was invested at 16% and the rest at 14%. After one year, the total interest earned was $13.74. How much did he invest at each percent?please answer FAST. I beg of you

Answer :

Let:

x = Investment in the first year

y = Investment in the second year

Since connor had 91.00 to invest:

[tex]x+y=91.00[/tex]

Using the simple interest formula:

For 16%:

[tex]\begin{gathered} I1=PV\cdot r\cdot t \\ so\colon \\ I1=x\cdot0.16\cdot1 \\ I1=0.16x \end{gathered}[/tex]

For 14%:

[tex]\begin{gathered} I2=PV\cdot r\cdot t \\ so\colon \\ I2=y\cdot0.14\cdot2 \\ I2=0.14y \end{gathered}[/tex]

After one year, the total interest earned was $13.74. so:

[tex]I1+I2=13.74=0.16x+0.14y[/tex]

So:

[tex]\begin{gathered} x+y=91_{\text{ }}(1) \\ 0.16x+0.14y=13.74_{\text{ }}(2) \end{gathered}[/tex]

From (1):

[tex]x=91-y_{\text{ }}(3)[/tex]

Replace (3) into (2):

[tex]\begin{gathered} 0.16(91-y)+0.14y=13.74 \\ 14.56-0.16y+0.14y=13.74 \\ -0.02y=-0.82 \\ y=41 \end{gathered}[/tex]

Replace the value of y into (3):

[tex]\begin{gathered} x=91-41 \\ x=50 \end{gathered}[/tex]

He invested $50 at 16% and $41 at 14%

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