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.