Jody invested $4400 less in account paying 4% simple interest than she did in an account paying 3 percent simple interest. At the end of the first year, the total interest from both accounts was $592. find the amount invested in each account

Respuesta :

The rule of the simple interest is

[tex]I=P\times R\times T[/tex]

P is the initial amount

R is the rate in decimal

T is the time

Assume that she invested $x in the account that paid 3% simple interest

then she invested x - 4400 dollars in the account that paid 4% simple interest

Then let us find each interest, then add them, equate the sum by 592

[tex]\begin{gathered} P1=x-4400 \\ R1=\frac{4}{100}=0.04 \\ T1=1 \\ I1=(x-4400)\times0.04\times1 \end{gathered}[/tex]

Let us simplify it

[tex]\begin{gathered} I1=0.04(x)-0.04(4400) \\ I1=0.04x-176 \end{gathered}[/tex][tex]\begin{gathered} P2=x \\ R2=\frac{3}{100}=0.03 \\ T2=1 \\ I2=x\times0.03\times1 \\ I2=0.03x \end{gathered}[/tex]

Since the total interest is $592, then

[tex]\begin{gathered} I1+I2=592 \\ 0.04x-176+0.03x=592 \end{gathered}[/tex]

Add the like terms on the left side

[tex]\begin{gathered} (0.04x+0.03x)-176=592 \\ 0.07x-176=592 \end{gathered}[/tex]

Add 176 to both sides

[tex]\begin{gathered} 0.07x-176+176=592+176 \\ 0.07x=768 \end{gathered}[/tex]

Divide both sides by 0.07 to find x

[tex]\begin{gathered} \frac{0.07x}{0.07}=\frac{768}{0.07} \\ x=10971.42857 \end{gathered}[/tex]

Then She invested about 10971 dollars in the account of 3%

Since 10971 - 4400 = 6571

Then she invested about