Last year, Linda had 30,000 to invest. She invested some of it in an account that paid 7% simple interest per year, and she invested the rest in an account that paid 6% simple I interest per year. After one year, she received a total of $1970 in interest. How much did she invest in each account?
Accepted Solution
A:
let x be the amount she invested in the 7% interest account let y be the amount she invested in the 6% interest account
assume linda spent all of her allocated interest money in investment, so x and y must add up to 30000
x + y = 30 000
interest equation:
interest = (investment )· (interest rate)· (time)
interest rate is the decimal percentage form. the time for both accounts is one year because the only info we're given as amount of total interest from both accounts after one year.
interest from the 7% account = x(0.07)(1) = 0.07x
interest from the 6% account = y(0.06)(1) = 0.06y
The total interest is 1970. Therefore
interest from the 7% account + interest from the 6% account = 1970 0.07x + 0.06y = 1970
system of equations
x + y = 30 000 ........... (I) 0.07x + 0.06y = 1970 ............ (II)
solve by elimination. multiply both sides of equation (I) by -0.07 we do this so that we can get rid of 0.07x in equation (II)
-0.07x - 0.07y = -2100 ........... (I')
add equations (II) and this altered equation (I') together
0.07x + 0.06y = 1970 ...... (II) + -0.07x - 0.07y = -2100 ...... (I') ----------------------------------- 0x - 0.01y = -130 y = -130 / -0.01 y = $13 000
use x + y = 30 000 to find x.
x + y = 30 000 x + 13 000 = 30 000 x = 30 000 - 13 000 x = $17000
since x is the amount she invested in the 7% interest account and y is the amount she invested in the 6% interest account
invested $17000 in the 7% account invested $13000 in the 6% account