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Chapter 0: Problem 22

You invested \(\$ 11,000\) in two accounts paying \(5 \%\) and \(8 \%\) annual interest. If the total interest earned for the year was \(\$ 730,\) how much was invested at each rate?

Short Answer

Expert verified
So, the amount invested at 5% is $5000 and the amount invested at 8% is $6000.

Step by step solution

01

Set up the equations

The problem can be translated into a system of linear equations. Let's denote the amount invested at 5% as \(x\) and at 8% as \(y\). From the above conditions, we can form the following two equations: \(x + y = 11000\) (the total investment is $11,000) and \(0.05x + 0.08y = 730\) (the total interest earned is $730).
02

Solve the system of equations

Solve this system of equations either by substitution method or elimination method. Here, let's use the substitution method. To solve for \(x\), make \(x\) the subject in first equation: \(x = 11000 - y\). Substituting \(x\) in the second equation we get: \(0.05(11000 - y) + 0.08y = 730\). After simplifying, this will yield an equation in terms of \(y\).
03

Simplify and find the value of y

Simplify to get \(y\): \(550 - 0.05y + 0.08y = 730 \) which yields \(0.03y = 180 \) upon simplification. Then solve for \(y\) to find \(y = 6000\).
04

Calculate the value of x

Substitute \(y = 6000\) in the first equation: \(x = 11000 - 6000\). Thus, \(x = 5000\).

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