section 3.4 motion, money, and mixture problems\n3. elizabeth invested $16,000 for one year, part at 6%…

section 3.4 motion, money, and mixture problems\n3. elizabeth invested $16,000 for one year, part at 6% interest and part at 9% interest. if she earned a total of $1320 in interest how much did she invest at each rate?

section 3.4 motion, money, and mixture problems\n3. elizabeth invested $16,000 for one year, part at 6% interest and part at 9% interest. if she earned a total of $1320 in interest how much did she invest at each rate?

Answer

Explanation:

Step1: Let the amount invested at 6% be $x$.

Then the amount invested at 9% is $(16000 - x)$.

Step2: Calculate the interest from each part.

The interest from the 6% - investment is $0.06x$, and the interest from the 9% - investment is $0.09(16000 - x)$.

Step3: Set up the interest - total equation.

The total interest is $1320$, so $0.06x+0.09(16000 - x)=1320$.

Step4: Expand and simplify the equation.

$0.06x + 1440-0.09x=1320$. Combining like - terms gives $- 0.03x=1320 - 1440=-120$.

Step5: Solve for $x$.

Dividing both sides by $-0.03$, we get $x=\frac{-120}{-0.03}=4000$.

Step6: Find the amount invested at 9%.

The amount invested at 9% is $16000 - 4000 = 12000$.

Answer:

She invested $4000$ at 6% and $12000$ at 9%.