Answer:
The amount invested at 5% was $39,000 and the amount invested at 7% was $11,000
Step-by-step explanation:
we know that
The simple interest formula is equal to
![I=P(rt)](https://tex.z-dn.net/?f=I%3DP%28rt%29)
where
I is the Final Interest Value
P is the Principal amount of money to be invested
r is the rate of interest
t is Number of Time Periods
Let
x-----> the amount invested at 5%
50,000-x -----> the amount invested at 7%
so
![t=1\ year\\ I=\$2,720\\r_1=0.05\\r_2=0.07\\P_1=\$x\\P_2=\$(50,000-x)](https://tex.z-dn.net/?f=t%3D1%5C%20year%5C%5C%20I%3D%5C%242%2C720%5C%5Cr_1%3D0.05%5C%5Cr_2%3D0.07%5C%5CP_1%3D%5C%24x%5C%5CP_2%3D%5C%24%2850%2C000-x%29)
substitute in the formula above
![2,720=x(0.05*1)+(50,000-x)(0.07*1)](https://tex.z-dn.net/?f=2%2C720%3Dx%280.05%2A1%29%2B%2850%2C000-x%29%280.07%2A1%29)
solve for x
![2,720=0.05x+3,500-0.07x](https://tex.z-dn.net/?f=2%2C720%3D0.05x%2B3%2C500-0.07x)
![0.07x-0.05x=3,500-2,720\\0.02x=780\\x=\$39,000](https://tex.z-dn.net/?f=0.07x-0.05x%3D3%2C500-2%2C720%5C%5C0.02x%3D780%5C%5Cx%3D%5C%2439%2C000)
![(50,000-x)=\$11,000](https://tex.z-dn.net/?f=%2850%2C000-x%29%3D%5C%2411%2C000)
therefore
The amount invested at 5% was $39,000 and the amount invested at 7% was $11,000