Question 30976
Let's first define three variables to keep track of the amounts invested at each rate.

x = amount invested at 5%
y = amount invested at 6%
z = amount invested at 7%

This problem has three unknown values - let's establish the relationships between these unknowns.

We know that the amount invested at 6% (y) is 1/4 the amount invested at 5% (x).  Therefore, we can say:
{{{y = (1/4)*x}}}

Then, whatever is not invested at 5% or 6% will be invested at 7%.  The total amount of money Joe is investing is $40,000.  So, we can say:
{{{z = 40000 - x - y}}} (to find the "rest of the money" invested at 7%)
 or...

{{{z = 40000 - x - (1/4)*x}}}  (substituting the y equation we created earlier)


We know his total annual interest income is $2,530.  To find the income, we just take the amounts invested at each rate and multiply them by their respective rates (i.e. since x is the amount invested at 5%, we find the interest income by multiplying 0.05x).  This means:

{{{0.05*x + 0.06*y + 0.07*z = 2530}}}

 or...

{{{0.05*x + 0.06*(1/4)*x + 0.07*(40000 - x - (1/4)*x) = 2530}}}

Now, solve for x:

{{{0.05*x + 0.015*x + 0.07(40000 - 1.25*x) = 2530}}}  [simplify equation]


{{{0.05*x + 0.015*x + 0.07(40000 - 1.25*x) = 2530}}}  [simplify equation]


{{{0.065*x + 2800 - 0.0875*x = 2530}}}  [more simplifying the equation]


{{{2800 - 0.0225*x = 2530}}}  [combine like terms]


{{{-0.0225*x = - 270}}}  [combine like terms]


{{{x = 12,000}}}  [divide both sides by -0.0225 to isolate x]


Going back to our original equations:


x = amount invested at 5%
y = amount invested at 6% = (1/4)x
z = amount invested at 7% = 40,000 - x - y


We have x = 12,000.  
y = (1/4)(12,000) = 3,000.
z = 40,000 - 12,000 - 3,000 = 25,000.