Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN: 9780134463216
Author: Robert F. Blitzer
Publisher: PEARSON
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**Investment and Interest Calculation**

Michelle invested $18,000. Part of it was invested at 3% annual simple interest, and the rest was invested at 4%. Her interest income for the first year was $650. How much did she invest at each rate? Write a system of linear equations and solve.

**Steps to Set Up and Solve the System of Equations:**

1. Let's denote the amount invested at 3% as \( x \) and the amount invested at 4% as \( y \).

2. The total amount invested is $18,000:
   \[
   x + y = 18000
   \]

3. The total interest earned from both investments is $650. Therefore, using the interest rates:
   \[
   0.03x + 0.04y = 650
   \]

Now, you have a system of linear equations:
\[
\begin{cases}
x + y = 18000 \\
0.03x + 0.04y = 650
\end{cases}
\]

**Solving the System of Equations:**

1. Solve the first equation for \( y \):
   \[
   y = 18000 - x
   \]

2. Substitute \( y = 18000 - x \) into the second equation:
   \[
   0.03x + 0.04(18000 - x) = 650
   \]

3. Simplify and solve for \( x \):
   \[
   0.03x + 720 - 0.04x = 650 \\
   -0.01x + 720 = 650 \\
   -0.01x = 650 - 720 \\
   -0.01x = -70 \\
   x = 7000
   \]

4. Substitute \( x = 7000 \) back into \( y = 18000 - x \):
   \[
   y = 18000 - 7000 \\
   y = 11000
   \]

Therefore, Michelle invested $7,000 at 3% and $11,000 at 4%.
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Transcribed Image Text:**Investment and Interest Calculation** Michelle invested $18,000. Part of it was invested at 3% annual simple interest, and the rest was invested at 4%. Her interest income for the first year was $650. How much did she invest at each rate? Write a system of linear equations and solve. **Steps to Set Up and Solve the System of Equations:** 1. Let's denote the amount invested at 3% as \( x \) and the amount invested at 4% as \( y \). 2. The total amount invested is $18,000: \[ x + y = 18000 \] 3. The total interest earned from both investments is $650. Therefore, using the interest rates: \[ 0.03x + 0.04y = 650 \] Now, you have a system of linear equations: \[ \begin{cases} x + y = 18000 \\ 0.03x + 0.04y = 650 \end{cases} \] **Solving the System of Equations:** 1. Solve the first equation for \( y \): \[ y = 18000 - x \] 2. Substitute \( y = 18000 - x \) into the second equation: \[ 0.03x + 0.04(18000 - x) = 650 \] 3. Simplify and solve for \( x \): \[ 0.03x + 720 - 0.04x = 650 \\ -0.01x + 720 = 650 \\ -0.01x = 650 - 720 \\ -0.01x = -70 \\ x = 7000 \] 4. Substitute \( x = 7000 \) back into \( y = 18000 - x \): \[ y = 18000 - 7000 \\ y = 11000 \] Therefore, Michelle invested $7,000 at 3% and $11,000 at 4%.
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