Algebra?? Variables.... Word problem?

You would make x be the amount invested at 10% and (8000-x) the part invested at 12%.

Set up the equation like:

.10x + .12(8000-x) = 900

.10x + 960 - .12x = 900

-.02 x = -60

x = 3000

8000 - x = 5000

a + b = $8000

0.1*(8000-a) + 0.12*(8000-b) = $900

simplified:

800 - 0.1a + 960 - 0.12b = 900

rearranged:

a = $8000 - b

substituted:

800 - 0.1(8000-b) + 960 - 0.12b = 900

simplified again:

800 - 800 + 0.1b + 960 - 0.12b = 900

0.1b - 0.12b = 900 - 960

-0.02b = -60

b = -60 / -0.02 = 3000

therefore $3000 at 12% ($360)

and remaining $5000 at 10% ($500)

which totals $860

or $3000 at 10% ($300)

and $5000 at 12% ($600)

which totals $900

hehe i got it!! man that was tough...

you use y as the part of 8000 invested at 10% interest and x is part of 8000 invested at 12% interest

Let x = amount invested at 10%

then 8000 - x = amount invested at 12%

0.10x + 0.12(8000 - x) = 900

-0.02x = -60

x = 3,000 at 10%

8000 - x = 5,000 at 12%

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from the subject, u can arise w/ one extra equation-- y=z. bc "the kit includes the comparable volume of Hazelnut coffee as French roast coffee." then u ought to be waiting to unravel the equations.

use x and y for the variables