To solve the problem, we need to determine the amounts invested at two different interest rates given a set of conditions. Let [tex]\( x \)[/tex] be the amount invested at [tex]\( 8\% \)[/tex] and [tex]\( y \)[/tex] be the amount invested at [tex]\( 5\% \)[/tex]. We are given the following conditions:
1. There is \[tex]$1100 more invested at \( 8\% \) than at \( 5\% \):
\[
x = y + 1100
\]
2. The total annual interest received from both investments is \$[/tex]530:
[tex]\[
0.08x + 0.05y = 530
\][/tex]
Now, let's solve the system of equations step-by-step using the method of addition.
Step 1: Substitute the expression for [tex]\( x \)[/tex] from the first equation into the second equation.
Given:
[tex]\[
x = y + 1100
\][/tex]
Substitute [tex]\( x \)[/tex] in the second equation:
[tex]\[
0.08(y + 1100) + 0.05y = 530
\][/tex]
Step 2: Distribute the [tex]\( 0.08 \)[/tex] in the equation.
[tex]\[
0.08y + 0.08 \times 1100 + 0.05y = 530
\][/tex]
This simplifies to:
[tex]\[
0.08y + 88 + 0.05y = 530
\][/tex]
Step 3: Combine like terms.
[tex]\[
(0.08y + 0.05y) + 88 = 530
\][/tex]
[tex]\[
0.13y + 88 = 530
\][/tex]
Step 4: Isolate [tex]\( y \)[/tex] by subtracting 88 from both sides.
[tex]\[
0.13y = 530 - 88
\][/tex]
[tex]\[
0.13y = 442
\][/tex]
Step 5: Solve for [tex]\( y \)[/tex] by dividing both sides by 0.13.
[tex]\[
y = \frac{442}{0.13}
\][/tex]
[tex]\[
y = 3400
\][/tex]
Step 6: Substitute the value of [tex]\( y \)[/tex] back into the first equation to find [tex]\( x \)[/tex].
[tex]\[
x = y + 1100
\][/tex]
[tex]\[
x = 3400 + 1100
\][/tex]
[tex]\[
x = 4500
\][/tex]
Therefore, the amounts invested are:
- [tex]\(\$4500\)[/tex] at [tex]\( 8\% \)[/tex]
- [tex]\(\$3400\)[/tex] at [tex]\( 5\% \)[/tex]
So the final answers are:
[tex]\[
\$4500 \text{ invested at } 8\%
\][/tex]
[tex]\[
\$3400 \text{ invested at } 5\%
\][/tex]
