Question

Money is invested at two rates of interest. One rate is 8% and the other is 5%. if there is $1400 more invested at 8% than at 5%, find the amount invested at each rate if the total annual interest received is $840. let x= amount invested at 8% and y= amount invested at 5%. then the system that models the problem is {x=y 14000.08x 0.05y=840. solve the system by using the method of addition.

Answer 1

To solve the problem, set up a system of equations based on the given information and solve using the method of addition.

Step-by-step explanation:

To solve this problem, we can set up a system of equations based on the given information. Let x be the amount invested at 8% and y be the amount invested at 5%. The first equation is x = y + 1400, which represents that there is $1400 more invested at 8% than at 5%. The second equation is 0.08x + 0.05y = 840, which represents the total annual interest received.

We can now solve the system of equations using the method of addition. Multiply the first equation by 0.08 to make the x terms cancel out when added to the second equation:

0.08(x) = 0.08(y + 1400)

0.08x = 0.08y + 112

Add this to the second equation:

0.08x + 0.05y = 840

(0.08y + 112) + 0.05y = 840

0.13y + 112 = 840

0.13y = 728

y = 5600

Substitute this value of y back into the first equation to find x:

x = 5600 + 1400

x = 7000

Therefore, $7000 is invested at 8% and $5600 is invested at 5%.