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Chapter 1: Problem 39

You invested \(\$ 7000\) in two accounts paying \(6 \%\) and \(8 \%\) annual interest. If the total interest earned for the year was \(\$ 520,\) how much was invested at each rate?

Short Answer

Expert verified
\$ 4000 was invested at an interest rate of 6\% and \$3000 was invested at an interest rate of 8\%.

Step by step solution

01

Formulate 1st equation

Form a linear equation from the total amount invested. It is given that the total investment is \$7000 which is split into two parts, \(x\) and \(y\). Therefore, the equation becomes \(x+y = 7000\).
02

Formulate 2nd equation

Similarly, form a linear equation from the given total interest. It is given that the total interest from both investments is \$520. According to the interest formula \(I = PRT\), where \(I\) is interest, \(P\) is principal or starting amount, \(R\) is the interest rate and \(T\) is the time period, the equation is \(0.06x + 0.08y = 520\).
03

Solve the system

To solve the system, you can use substitution or elimination method. In this case, the substitution method can make things easier. First, from the equation \(x + y = 7000\), you can express \(x\) as \(x = 7000 - y\). Substitute this into the second equation to find value of \(y\).
04

Calculate the value of y

Now, replace \(x\) in the 2nd equation with the equation \(x = 7000 - y\). The equation becomes \(0.06 * (7000 - y) + 0.08y = 520\) . After solving, the amount invested at 8\% is \(y = \$ 3000\).
05

Calculate the value of x

Substitute the value of \(y\) (\$ 3000) into the first equation \(x + y = 7000\), to find the amount invested at 6\%, which is \(x = \$ 7000 - \$ 3000 = \$ 4000\)

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