Connie invested all of her $3000 savings,part at an annual rate of 6% and the rest at an annual rate of 9%.?

Question

Her annual income from both the investments was 8% of her total investment. How much did she invest at 6%?

Answer 1

x + y = 3000

y = 3000 - x

x * .06 + y * .09 = 3000 * .09

x * .06 + (3000 - x) * .08 = 240

x * .06 + 270 - x * .09 = 240

x (.06 - .09) = -30

x (-.03) = -30

x = -30/-.03 = 1000

x = 1000 @ 6%

Answer 2

0.06x+0.09y = 0.08*3000
x+y=3000
y=3000-x
0.06x +270-0.09x=240
so
0.03x=30 and x=$1000 so y= $2000
(x invested at6% and y invested at9%)