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A student invests 8% and 2100 less than that amount at 5% the investments produced a total of 233$ interest in 1 yr. How much did he invest at each rate?

I need help with percentage problems

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Mitch H. | Tutor in Natural Sciences & Math for University to Grades 6-12Tutor in Natural Sciences & Math for Uni...
I am going to restate what I believe you are trying to say, because it's not quite clear to me.
There are two investment amounts, P1 and P2.
P1 is invested at an 8% yield.
P2 is invested at $2100 less than the amount P1 at a 5% yield.
Therefore, P1 = P2 - $2100
In one year, the yields from both P1 and P2 sum up to $233.
The question asks you to determine (solve) P1 and P2.
First we need to know what the rate means.  Is 8% a value that is 8% per year, per month, per day?  When the period is not given in the financial world, the assumption is that the rate is a per year (annual) period.
Since we have two investments, P1 and P2, we will have two yields ("yield" is synonymous with "interest" here).
The basic formula for the amount of simple interest (I) earned on an amount representing the principal (P) at a periodic rate (r) for a certain number of periods (t) is:
I = Prt
Thus for P1, the interest earned is I1 = (P1)(r1)(t1).
We know that rate (r1) = 8% per year, and we know that period (t1) = 1 year.  We don't know P1 (we are trying to solve it).  We thus don't know I1 (yet).
Thus I1 = (P1)(0.08/yr)(1 yr) = (0.08)(P1)
Note that 8% = 8/100 = 0.08.  The dimension (or units) for P1 will be dollars ($), and since the product on the right side of the equation has the rate at "per year" (1/year) and the time is in years) that will cancel out the time dimensions, which we would expect.
For P2, the interest earned is I2 = (P2)(r2)(t2).
The rate (r2) = 5% = 5/100 = 0.05, and the period (t2) is also 1 year.
Thus I2 = (P2)(0.05/yr)(1 yr) = (0.05)(P2)
We also know that sum of the interest (I1 + I2) is $233.
When you count up the number of unknowns, it is four:  P1, P2, I1, and I2.
So let's see what we have now in terms of equations:
1.  P1 = P2 - $2100
2.  I1 = (0.08)P1
3.  I2 = (0.05)P2
4.  I1 + I2 = $233
That's good...we have four unknowns and we needed four equations.  The rest should be algebra.
Let's solve for P1, the one investment.  Using equation 1 we see P1 in terms of P2.  Let us get a value for P2 from equation 3 by dividing equation 3 by 0.05 on both sides:
A.  P1 = (I2/0.05) - $2100
Okay, now we need to substitute I2, which we can get from re-arranging equation 4:  I2 = $233 - I1
B.  P1 = ([$233 - I1]/0.05) - $2100
Finally we have to substitute for I1, which we get from equation 2:
C.  P1 = ([$233 - (0.08)P1)/0.05 - $2100
This is good, because we have the equation in terms of the unknown now, and we only need to do more algebra to get P1 on one side of the equation, right?  When I did the algebra, it looked like this:
D.  P1 = [$233 - (0.05)($2100)] / (0.08 + 0.05)
This evaluates to P1 = $984.62.
Since P2 = P1 + $2100 (from rearranging equation 1), then P2 = $3084.62.
Does this make sense?
From equation 2, the interest on P1 should be I1 = (0.08)($984.62)  = $78.77
From equation 3, the interest on P2 should be I2 = (0.05)($3084.62) = $154.23
When you add I1 and I2, you get $154.23 + $78.77 = $233.00
Christopher D. | Mathematician and ProgrammerMathematician and Programmer
4.9 4.9 (299 lesson ratings) (299)
0.08 * x - 2100 = 0.05 * x
0.03 * x = 2100
x = 70000
0.08 * x = 5600
0.05 * x = 3500