HELP! Please

Calvin invested $7500 for one year, part 12% annual interest and the rest at 10% annual interest. His total interest for the year was $890. How much money did he invest at 12%? Show work.

$X @ 12%.

$(7500-X) @ 10%

I = 0.12X + 0.1(7500-X) = $890.
0.12x + 750 - 0.1x = 890.
02x = 890 - 750 = 140
X = $7,000 @ 12%.

To find the amount Calvin invested at 12%, let's assume he invested x amount of money at 12%.

Since Calvin invested a total amount of $7500, the amount he invested at 10% can be calculated as the difference between $7500 and x.

The calculation for the interest earned at 12% can be derived as follows:

Interest at 12% = (x * 12%)
Interest at 10% = ((7500 - x) * 10%)

We know that the total interest earned for the year is $890, so we can set up the equation:

(x * 12%) + ((7500 - x) * 10%) = $890

Now, we can solve this equation to find the value of x:

0.12x + 0.10(7500 - x) = 890
0.12x + 750 - 0.10x = 890
0.02x + 750 = 890
0.02x = 890 - 750
0.02x = 140
x = 140 / 0.02
x = 7000

Therefore, Calvin invested $7000 at 12% interest.