A total of $5000 is invested at 2%, 3% and 4%. The amount invested at 4% equals the total amount invested at 2% and 3%. The total interest for one years is $145. If possible, find the amount invested in each interest rate. Interpret your answer.

Let x=the amount invested at 2%

y=the amount invested at 3%, and

z=the amount invested at 4%,

We know the total of all three investments is $5000, so

x+y+z=5000

The amount invested at 4%, z, equals the total amount invested at 2% and 3%, so

z=x+y

Assume the interest is compounded annually, then

0.02x+0.03y+0.04z=145

We now have three equations and three unknowns!

x+y+z=5000 [1]

z=x+y [2]

0.02x+0.03y+0.04z=145 [3]

First eliminate z by substituting [2] into [1] then [2] into [3]:

x+y+(x+y)=5000

2x+2y=5000, so

x+y=2500 [4]

0.02x+0.03y+0.04(x+y)=145 [5]

0.06x+0.07y=145

We now have two equations and two unknowns, x and y:

x+y=2500 [6]

0.06x+0.07y=145 [7]

From [6], y=2500-x. Substitute this into [7]:

0.06x+0.07(2500-x)=145

0.06x+175-0.07x=145

-0.01x+175=145

-0.01x=145-175=-30

x=30/0.01=3000

y=2500-x=2500-3000=-500

z=x+y=3000-500=2500

Since a negative investment, y, makes no sense, my assumption about the compounding period could be wrong.

I showed all steps so those a little weak with algebra can follow the solution.