Janet invested $26,000, part at 6% and part at 3%. If the total interest at the end of the year is $1,080, how much did she invest at 6%?

Let x = amount invested @ 6%

then 26,000-x = amount invested @ 3%.

so 0.06 x = interest from the 6% money.
0.03(26,000-x) = interest from the 3% money.
Just add them together, make them equal to 1080 and solve for x. Post your work if you get stuck.

9,000

To find out how much Janet invested at 6%, we can set up an algebraic equation based on the given information.

Let's assume Janet invested an amount, x, at 6% interest. So, the amount she invested at 3% interest would be (26,000 - x), as she invested a total of $26,000.

Now, we need to calculate the interest earned on each investment.

The interest earned on the amount invested at 6% is given by x * 6% = 0.06x.
The interest earned on the amount invested at 3% is given by (26,000 - x) * 3% = 0.03(26,000 - x).

Since the total interest at the end of the year is $1,080, we can set up the equation:

0.06x + 0.03(26,000 - x) = 1,080.

Now, let's solve this equation to find the value of x, which represents the amount that Janet invested at 6%.

0.06x + 0.03(26,000 - x) = 1,080.

Simplifying the equation:

0.06x + 780 - 0.03x = 1,080.

Combining like terms:

0.03x + 780 = 1,080.

Subtracting 780 from both sides:

0.03x = 300.

Now, divide both sides of the equation by 0.03:

x = 300 / 0.03 = 10,000.

Therefore, Janet invested $10,000 at 6%.