you invested $20,000 in two accounts paying 7% and 9% annual interest, respectively. if the total interest…

you invested $20,000 in two accounts paying 7% and 9% annual interest, respectively. if the total interest earned for the year was $1760, how much was invested at each rate? the amount invested at 7% is $ . the amount invested at 9% is $ .

you invested $20,000 in two accounts paying 7% and 9% annual interest, respectively. if the total interest earned for the year was $1760, how much was invested at each rate? the amount invested at 7% is $ . the amount invested at 9% is $ .

Answer

Explanation:

Step1: Let the variables

Let $x$ be the amount invested at 7% and $y$ be the amount invested at 9%. We know that $x + y=20000$ (total - investment equation), so $y = 20000 - x$. The interest - earned equation is $0.07x+0.09y = 1760$.

Step2: Substitute $y$ into the interest - earned equation

Substitute $y = 20000 - x$ into $0.07x + 0.09y=1760$. We get $0.07x+0.09(20000 - x)=1760$. Expand the equation: $0.07x+1800 - 0.09x=1760$.

Step3: Combine like terms

Combine the $x$ terms: $0.07x-0.09x=1760 - 1800$. $-0.02x=-40$.

Step4: Solve for $x$

Divide both sides of the equation by $- 0.02$: $x=\frac{-40}{-0.02}=2000$.

Step5: Solve for $y$

Since $y = 20000 - x$, substitute $x = 2000$ into this equation. So $y=20000 - 2000 = 18000$.

Answer:

The amount invested at 7% is $2000$. The amount invested at 9% is $18000$.