Solve the following word problem.
Money is invested at two rates of interest. One rate is 7%
and the other is 5%
. If there is $1000
more invested at 7%
than at 5%
, find the amount invested at each rate if the total annual interest received is $790
. Let x=
amount invested at 7%
and y=
amount invested at 5%
. Then the system that models the problem is {x=y+1000 0.07x+0.05y=790
. Solve the system by using the method of addition.
To solve the system using the method of addition, we can eliminate one variable by adding the two equations together.
Since the first equation is x = y + 1000, we can substitute this expression for x in the second equation.
0.07(y + 1000) + 0.05y = 790
Expanding and simplifying:
0.07y + 70 + 0.05y = 790
0.12y + 70 = 790
Subtracting 70 from both sides:
0.12y = 720
Dividing both sides by 0.12:
y = 720 / 0.12
y = 6000
Now, substitute this value of y back into the first equation to solve for x:
x = y + 1000
x = 6000 + 1000
x = 7000
So, the amount invested at a 7% interest rate is $7000 and the amount invested at a 5% interest rate is $6000.