algebra word problems

Question

larry mitchell invested part of his 27000, advance to 3% annual simple interest and the rest at 9% annual simple interest. If his total yearly interest from both accounts was $1,770 find the amount invested at each rate

Answer 1

Let T = amount invested at 3% interest

Let N = amount invested at 9% interest

 

27000 total  ==>  T + N = 27000  ==>  T = 27000 - N

 

Total yearly interest was 1770  ==>  .03*T + .09*N  

 

Substitute the value for T in to the 2nd equation 

 

.03*(27000 - N) + .09*N = 1770

810 - .03*N + .09*N = 1770

.06*N = 960

N = 960 / .06 = 16000

 

So T = 27000 - 16000 = 11000

 

$11000 was invested at 3% and $16000 was invested at 9%

 

Check:  11000 + 16000 = 27000

            .03*11000 + .09*16000 = 330 + 1440 = 1770