Question

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.

Answer 1

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.