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put this solution on YOUR website!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%.
