Question

Two investments are made totalling 8800 Pesos. For a certain year, these investments yield 1326 Pesos in simple interest. Part of the 8800 is invested at 16% and part at 14%. Find the amount invested at each rate ???

Answer 1

x*0.16+(8800-x)*0.14 = 1326 ... solve for x

Answer 2

Let x=Amount invested at 16%
Let y=Amount invested at 14%

Now you set up two equations:

0.16x+0.14y=1326
x+y=8800 Multiply this equation by 0.14 to eliminate y and find x

0.16x+0.14y=1326
0.14x+0.14y=1232 Subtract this equation from the first one to find x

0.02x=94
x=4700 invested at 16%

x+y=8800
y=8800-x
y=8800-4700
y=4100 invested at 14%