Question 260828
total invested is 8,000


let x be one part.
let y be the other part.


you have x + y = 8000 (equation 1)


total annual income is 860.000


x part is invested at 10%
y part is invested at 12%


you have .10*x + .12*y = 860 (equation 2)


you need to solve these 2 equations simultaneously.


solve for y in equation 1 to get y = 8000 - x


substitute for y in equation 2 to get .10*x + .12*(8000-x) = 860


solve for x in equation 2.


equation is:


.10*x + .12*(8000-x) = 860


simplify to get:


.10*x + .12*8000 - .12*x = 860


simplify to get:


-.02*x + 960 = 860


subtract 960 from both sides of this equation to get:


-.02*x = -100


divide both sides of this equation by -.02 to get:


x = 5000


that means y = 3000 because x + y = 8000.


you have x = 5000 and y = 3000.


substitute in equation 2 to get:


.10*5000 + .12*3000 = 860


simplify to get:


500 + 360 = 860 which is true confirming the values for x and y are good.


your answer is:


$5000 was invested at 10% and $3000 was invested at 12%.