Solving Applied Problems with 2 Equations

Question

$5900 is invested, part of it at 12% and part of it at 6%. For a certain year, the total yield is $540.00. How much was invested at each rate?

Answer 1

Hi Emily,

 

Let x = amount invested at 12%

let y = amount invested at 6%

 

then

 

x + y = 5900

.12x + .06y = 540

 

now multiply the the first equatin by .06

 

.06x + .06y = 354

.12x + .06y = 540

 

now subtract

 

-.06x = -186

 

x = 3100            (amount invested at 12%

 

y = 5900 - 3100

y = 2800            (amount invested at 6%

 

Hope this helps