Answer :
To find the amount of money invested at each rate, let's denote the variables as follows:
- [tex]\( x \)[/tex] as the amount invested at [tex]\( 4\% \)[/tex].
- [tex]\( y \)[/tex] as the amount invested at [tex]\( 11\% \)[/tex].
- [tex]\( z \)[/tex] as the amount invested at [tex]\( 15\% \)[/tex].
From the problem, we have the following information:
1. The total investment is \[tex]$7200. 2. The total annual income from these investments is \$[/tex]872.
3. The amount of money invested at [tex]\( 15\% \)[/tex] was \[tex]$1000 more than the amounts invested at \( 4\% \) and \( 11\% \) combined. We can set up the following system of equations based on the information given: 1. \( x + y + z = 7200 \) (total investment) 2. \( 0.04x + 0.11y + 0.15z = 872 \) (total annual income) 3. \( z = 1000 + x + y \) (the amount invested at \( 15\% \) was $[/tex]1000 more than the combined amounts invested at [tex]\( 4\% \)[/tex] and [tex]\( 11\% \)[/tex])
Now, we can solve the system of equations step-by-step.
Step 1: Substitute equation (3) into equation (1).
We have [tex]\( z = 1000 + x + y \)[/tex]. Substitute this into [tex]\( x + y + z = 7200 \)[/tex]:
[tex]\[ x + y + (1000 + x + y) = 7200 \][/tex]
[tex]\[ x + y + 1000 + x + y = 7200 \][/tex]
[tex]\[ 2x + 2y + 1000 = 7200 \][/tex]
[tex]\[ 2x + 2y = 6200 \][/tex]
[tex]\[ x + y = 3100 \quad \text{(simplified equation)} \][/tex]
Step 2: Solve for [tex]\( z \)[/tex] using equation (1).
Since [tex]\( x + y = 3100 \)[/tex]:
[tex]\[ z = 1000 + x + y \][/tex]
[tex]\[ z = 1000 + 3100 \][/tex]
[tex]\[ z = 4100 \][/tex]
Step 3: Substitute the values of [tex]\( z \)[/tex] into equation (2) to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
We know [tex]\( 0.04x + 0.11y + 0.15z = 872 \)[/tex] and [tex]\( z = 4100 \)[/tex]:
[tex]\[ 0.04x + 0.11y + 0.15(4100) = 872 \][/tex]
[tex]\[ 0.04x + 0.11y + 615 = 872 \][/tex]
[tex]\[ 0.04x + 0.11y = 257 \][/tex]
Given [tex]\( x + y = 3100 \)[/tex], we solve the two equations simultaneously:
1. [tex]\( x + y = 3100 \)[/tex]
2. [tex]\( 0.04x + 0.11y = 257 \)[/tex]
Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from equation (1):
[tex]\[ y = 3100 - x \][/tex]
Substitute [tex]\( y \)[/tex] into equation (2):
[tex]\[ 0.04x + 0.11(3100 - x) = 257 \][/tex]
[tex]\[ 0.04x + 341 - 0.11x = 257 \][/tex]
[tex]\[ 0.04x - 0.11x + 341 = 257 \][/tex]
[tex]\[ -0.07x = -84 \][/tex]
[tex]\[ x = 1200 \][/tex]
Now, substitute [tex]\( x = 1200 \)[/tex] into [tex]\( y = 3100 - x \)[/tex]:
[tex]\[ y = 3100 - 1200 \][/tex]
[tex]\[ y = 1900 \][/tex]
Finally, we verify that the amounts satisfy all conditions. Thus, the amounts invested are:
- [tex]\(\$1200\)[/tex] at [tex]\(4\%\)[/tex]
- [tex]\(\$1900\)[/tex] at [tex]\(11\%\)[/tex]
- [tex]\(\$4100\)[/tex] at [tex]\(15\%\)[/tex]
Therefore, the person invested [tex]\(\$1200\)[/tex] at [tex]\(4\%\)[/tex], [tex]\(\$1900\)[/tex] at [tex]\(11\%\)[/tex], and [tex]\(\$4100\)[/tex] at [tex]\(15\%\)[/tex].
- [tex]\( x \)[/tex] as the amount invested at [tex]\( 4\% \)[/tex].
- [tex]\( y \)[/tex] as the amount invested at [tex]\( 11\% \)[/tex].
- [tex]\( z \)[/tex] as the amount invested at [tex]\( 15\% \)[/tex].
From the problem, we have the following information:
1. The total investment is \[tex]$7200. 2. The total annual income from these investments is \$[/tex]872.
3. The amount of money invested at [tex]\( 15\% \)[/tex] was \[tex]$1000 more than the amounts invested at \( 4\% \) and \( 11\% \) combined. We can set up the following system of equations based on the information given: 1. \( x + y + z = 7200 \) (total investment) 2. \( 0.04x + 0.11y + 0.15z = 872 \) (total annual income) 3. \( z = 1000 + x + y \) (the amount invested at \( 15\% \) was $[/tex]1000 more than the combined amounts invested at [tex]\( 4\% \)[/tex] and [tex]\( 11\% \)[/tex])
Now, we can solve the system of equations step-by-step.
Step 1: Substitute equation (3) into equation (1).
We have [tex]\( z = 1000 + x + y \)[/tex]. Substitute this into [tex]\( x + y + z = 7200 \)[/tex]:
[tex]\[ x + y + (1000 + x + y) = 7200 \][/tex]
[tex]\[ x + y + 1000 + x + y = 7200 \][/tex]
[tex]\[ 2x + 2y + 1000 = 7200 \][/tex]
[tex]\[ 2x + 2y = 6200 \][/tex]
[tex]\[ x + y = 3100 \quad \text{(simplified equation)} \][/tex]
Step 2: Solve for [tex]\( z \)[/tex] using equation (1).
Since [tex]\( x + y = 3100 \)[/tex]:
[tex]\[ z = 1000 + x + y \][/tex]
[tex]\[ z = 1000 + 3100 \][/tex]
[tex]\[ z = 4100 \][/tex]
Step 3: Substitute the values of [tex]\( z \)[/tex] into equation (2) to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex].
We know [tex]\( 0.04x + 0.11y + 0.15z = 872 \)[/tex] and [tex]\( z = 4100 \)[/tex]:
[tex]\[ 0.04x + 0.11y + 0.15(4100) = 872 \][/tex]
[tex]\[ 0.04x + 0.11y + 615 = 872 \][/tex]
[tex]\[ 0.04x + 0.11y = 257 \][/tex]
Given [tex]\( x + y = 3100 \)[/tex], we solve the two equations simultaneously:
1. [tex]\( x + y = 3100 \)[/tex]
2. [tex]\( 0.04x + 0.11y = 257 \)[/tex]
Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from equation (1):
[tex]\[ y = 3100 - x \][/tex]
Substitute [tex]\( y \)[/tex] into equation (2):
[tex]\[ 0.04x + 0.11(3100 - x) = 257 \][/tex]
[tex]\[ 0.04x + 341 - 0.11x = 257 \][/tex]
[tex]\[ 0.04x - 0.11x + 341 = 257 \][/tex]
[tex]\[ -0.07x = -84 \][/tex]
[tex]\[ x = 1200 \][/tex]
Now, substitute [tex]\( x = 1200 \)[/tex] into [tex]\( y = 3100 - x \)[/tex]:
[tex]\[ y = 3100 - 1200 \][/tex]
[tex]\[ y = 1900 \][/tex]
Finally, we verify that the amounts satisfy all conditions. Thus, the amounts invested are:
- [tex]\(\$1200\)[/tex] at [tex]\(4\%\)[/tex]
- [tex]\(\$1900\)[/tex] at [tex]\(11\%\)[/tex]
- [tex]\(\$4100\)[/tex] at [tex]\(15\%\)[/tex]
Therefore, the person invested [tex]\(\$1200\)[/tex] at [tex]\(4\%\)[/tex], [tex]\(\$1900\)[/tex] at [tex]\(11\%\)[/tex], and [tex]\(\$4100\)[/tex] at [tex]\(15\%\)[/tex].
