To solve the problem of determining how much was invested at each interest rate, let's break it down step-by-step.
### Step-by-Step Solution
1. Define the variables:
- Let [tex]\( x \)[/tex] be the amount invested at [tex]\( 6\% \)[/tex].
- Let [tex]\( y \)[/tex] be the amount invested at [tex]\( 8\% \)[/tex].
2. Set up the system of equations:
- The total amount invested is \[tex]$7000:
\[ x + y = 7000 \]
- The total interest earned from both investments is \$[/tex]520:
[tex]\[ 0.06x + 0.08y = 520 \][/tex]
3. Solve the system of equations:
First Equation:
[tex]\[ x + y = 7000 \][/tex]
Second Equation:
[tex]\[ 0.06x + 0.08y = 520 \][/tex]
4. From the first equation, express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]:
[tex]\[ y = 7000 - x \][/tex]
5. Substitute [tex]\( y \)[/tex] into the second equation:
[tex]\[ 0.06x + 0.08(7000 - x) = 520 \][/tex]
6. Simplify and solve for [tex]\( x \)[/tex]:
[tex]\[ 0.06x + 0.08 \cdot 7000 - 0.08x = 520 \][/tex]
[tex]\[ 0.06x + 560 - 0.08x = 520 \][/tex]
[tex]\[ -0.02x + 560 = 520 \][/tex]
[tex]\[ -0.02x = 520 - 560 \][/tex]
[tex]\[ -0.02x = -40 \][/tex]
[tex]\[ x = \frac{-40}{-0.02} \][/tex]
[tex]\[ x = 2000 \][/tex]
7. Now substitute [tex]\( x \)[/tex] back into the expression for [tex]\( y \)[/tex]:
[tex]\[ y = 7000 - x \][/tex]
[tex]\[ y = 7000 - 2000 \][/tex]
[tex]\[ y = 5000 \][/tex]
### Conclusion
- The amount invested at [tex]\( 6\% \)[/tex] is [tex]\( \$2000 \)[/tex].
- The amount invested at [tex]\( 8\% \)[/tex] is [tex]\( \$5000 \)[/tex].
Therefore, you invested \[tex]$2000 at a \( 6\% \) interest rate and \$[/tex]5000 at an [tex]\( 8\% \)[/tex] interest rate.
