1. Investment Problems

2. We’re asked to find the amount invested at each rate, so …
A college student earned $5000 during summer vacation working as a waiter in a popular restaurant. The student invested part of the money at 8% and the rest at 6%. If the student received a total of $380 in interest at the end of the year, how much was invested at 8%?
We’re asked to find the amount invested at each rate, so …
Let x = the amount invested at 8%
Let y = the amount invested at 6%

3. Our first equation is x + y = 5000
Now we need to find out two relationships between these amounts:
1) A college student earned $5000 during summer vacation .
The amount invested at 8% + the amount invested at 6% = 5000
Our first equation is x + y = 5000

4. Our second equation is .08x + .06y = 380
We need one more relationship between the amounts:
2) The student received a total of $380 in interest at the end of the year.
Interest = Principal (amount invested) x rate x time
Interest at 8% = amount invested x .08 x 1 year
Interest at 8% = .08x
Interest at 6% = .06y
Our second equation is .08x + .06y = 380

5. Now we have two equations in two unknowns
x + y = 5000
.08x + .06y = 380
I would solve this by elimination
.08x + .06y = multiply by 100 to clear the decimals
8x + 6y = 38000

6. Multiply the top equation by -6 to clear the y column:
x + y = 5000
8x + 6y = 38000
-6x - 6y =
Now just add the equations together. The y’s will add up to zero and disappear!
2x = 8000
x = 4000

7. Now let’s find the amount invested at 6% and check our work
x + y = 5000
x = 4000
y = 5000
y = 1000
Now we have $4000 at 8% and $1000 at 6%. Does this fit our story?
Does $4000 at 8% + $1000 at 6% = 380
4000(.08) (.06)
= 380
We have a winner!
Be sure you answer the question!
How much was invested at 8%? $4000 was invested at 8%