nico is saving money for his college education. He invests some money at 6%, and $1400 less that that amount at 5%. The investments produced a total of $172 interest in 1 yr. how much did he invest at each rate?

Let x be the amount he invested at 6%.

Then x-1400 is the amount he invested at 5%.

Total interest after 1 year
x*0.06+(x-1400)*0.05=172
0.11x=172+1400*0.05
0.11x=242
x=2200

Check
2200*0.06+800*0.05=132+40=172 ok.

To find out how much Nico invested at each rate, we can set up a system of equations based on the information given.

Let's say Nico invested an amount x at 6% and $1400 less than that amount (x - $1400) at 5%.

The interest earned from the investment at 6% is calculated as 0.06x.
The interest earned from the investment at 5% is calculated as 0.05(x - $1400).

The total interest earned from both investments is $172.

Therefore, we can write the equation:
0.06x + 0.05(x - $1400) = $172

Now we can solve for x:

0.06x + 0.05x - 0.05($1400) = $172
0.06x + 0.05x - $70 = $172
0.11x - $70 = $172

Adding $70 to both sides:
0.11x = $242

Now divide by 0.11 to isolate x:
x = $242 / 0.11
x ≈ $2,200

So, Nico invested approximately $2,200 at a rate of 6%.

To find out how much he invested at 5%, we can substitute the value of x back into our expression (x - $1400):
x - $1400 = $2,200 - $1400
x - $1400 ≈ $800

Therefore, Nico invested approximately $800 at a rate of 5%.