Ms. Buxx invested a total of $2000 in two saving accounts. The first account pays 3% intrest per year. The second accout pays 5% intrest per year. If the intrest form both accounts totals $84 dollars per year how much is invested in each account?

Solve these two equations in two unknowns. Let X be the amount invested at 3% and Y be the amount invested at 5%.

X + Y = 2000
0.03 X + 0.05 Y = 84

I recommend substituting 2000-X for Y in the second equaion, and than solving for X

The first step will give you
0.03 X + 0.05(2000-X) = 0
0.03 X + 100 - 0.05X = 0

Take it from there. Solve for X and then use Y = 2000 - X

But then you get x=5000. Which is much more than the investment.

My mistake, sorry.

0.03 X + 100 -0.05X = 84
16 = 0.02 X
X = 800
Y = 1200

really and truly it would be simpler if you multiply everything in the second equation by 100 then solve with whole numbers

To solve this problem, let's assign variables to the amounts invested in each account. Let's call the amount invested in the first account "x" and the amount invested in the second account "2000 - x" (since the total invested is $2000).

Now, we can use the interest rates and the total interest earned to set up two equations.

The first equation represents the interest from the first account:

(x) * (3%) = 0.03x

The second equation represents the interest from the second account:

(2000 - x) * (5%) = 0.05(2000 - x)

Since we know that the total interest earned is $84, we can set up the equation:

0.03x + 0.05(2000 - x) = 84

Now, let's solve the equation to find the value of x:

0.03x + 0.05(2000 - x) = 84
0.03x + 100 - 0.05x = 84
-0.02x = -16
x = -16 / (-0.02)
x = 800

Therefore, $800 is invested in the first account (at 3% interest), and $1200 is invested in the second account (at 5% interest).