Question 1178756: A natural history museum borrows $2,000,000 at simple annual interest to purchase new exhibits. Some of the money is borrowed at 5%, some at 6.5%, and some at 7.5%. Use a system of linear equations to determine how much is borrowed at each rate given that the total annual interest is $121,250 and the amount borrowed at 6.5% is four times the amount borrowed at 7.5%. Solve the system of linear equations using matrices.
at 5%
at 6.5%
at 6.5%
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's set up the system of linear equations and solve it using matrices.
**1. Define the Variables:**
* x = amount borrowed at 5%
* y = amount borrowed at 6.5%
* z = amount borrowed at 7.5%
**2. Formulate the Equations:**
* **Total Amount Borrowed:** x + y + z = 2,000,000
* **Total Annual Interest:** 0.05x + 0.065y + 0.075z = 121,250
* **Relationship Between y and z:** y = 4z
**3. Rewrite the Equations in Standard Form:**
1. x + y + z = 2,000,000
2. 0.05x + 0.065y + 0.075z = 121,250
3. 0x + y - 4z = 0
To simplify the second equation, multiply it by 1000:
1. x + y + z = 2,000,000
2. 50x + 65y + 75z = 121,250,000
3. y - 4z = 0
**4. Represent the System as an Augmented Matrix:**
```
[ 1 1 1 | 2000000 ]
[ 50 65 75 | 121250000 ]
[ 0 1 -4 | 0 ]
```
**5. Solve Using Row Operations:**
1. **Eliminate x from the second row:**
* R2 = R2 - 50 * R1
```
[ 1 1 1 | 2000000 ]
[ 0 15 25 | 21250000 ]
[ 0 1 -4 | 0 ]
```
2. **Swap R2 and R3 to make R2 easier to work with:**
```
[ 1 1 1 | 2000000 ]
[ 0 1 -4 | 0 ]
[ 0 15 25 | 21250000 ]
```
3. **Eliminate y from the third row:**
* R3 = R3 - 15 * R2
```
[ 1 1 1 | 2000000 ]
[ 0 1 -4 | 0 ]
[ 0 0 85 | 21250000 ]
```
4. **Solve for z:**
* 85z = 21,250,000
* z = 21,250,000 / 85
* z = 250,000
5. **Solve for y:**
* y - 4z = 0
* y = 4z
* y = 4 * 250,000
* y = 1,000,000
6. **Solve for x:**
* x + y + z = 2,000,000
* x = 2,000,000 - y - z
* x = 2,000,000 - 1,000,000 - 250,000
* x = 750,000
**6. Answers:**
* Amount borrowed at 5% (x): $750,000
* Amount borrowed at 6.5% (y): $1,000,000
* Amount borrowed at 7.5% (z): $250,000
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