Belinda had $20,000 to invest. She invested part of it at 10% and the remainder at 12%. If

her income from the two investments was $2160, then how much did she invest at each rate?

Let x be the amount invested at 10% and y be the amount invested at 12%. We can

summarize all of the given information in a table:
We can write one equation about the amounts invested and another about the interest
from the investments:
x + y = 20,000 Total amount invested
0.10x + 0.12y = 2160 Total interest
Solve the first equation for x to get x = 20,000 - y. Substitute 20,000 - y for x in
the second equation:
0.10x + 0.12y = 2160
0.10(20,000 - y) = 0.12y � 2160 Replace x by 20,000 - y.
2000 - 0.10y + 0.12y = 2160 Solve for y.
0.02y = 160
y = 8000
x = 12,000 Because x � 20,000 � y
To check this answer, find 10% of $12,000 and 12% of $8000:
0.10(12,000) = 1,200
0.12(8000) = 960
Because $1200 + $960 = $2160 and $8000 + $12,000 = $20,000, we can be certain
that Belinda invested $12,000 at 10% and $8000 at 12%.

To solve this problem, let's define the following variables:

Let x be the amount Belinda invested at 10%.
Then, the remainder (20000 - x) represents the amount she invested at 12%.

Now, we can set up an equation based on the given information.

The income from the investment at 10% is calculated as 10% of x, which is 0.10x.
The income from the investment at 12% is calculated as 12% of (20000 - x), which is 0.12(20000 - x).
The total income from both investments is given as $2160.

So our equation is: 0.10x + 0.12(20000 - x) = 2160.

To solve this equation, we can simplify and solve for x:

0.10x + 0.12(20000 - x) = 2160
0.10x + 2400 - 0.12x = 2160
0.10x - 0.12x = 2160 - 2400
-0.02x = -240
x = (-240) / (-0.02)
x = 12000

Therefore, Belinda invested $12,000 at 10% and the remainder, $8,000, at 12%.