Question 515404
Let N be the original number of nickels, and let Q be the original number of quarters.  N nickels are worth 5N cents, and Q quarters are worth 25Q cents, so N nickels and Q quarters are worth 5N + 25Q cents.  We know that the original number of nickels and quarters are worth $7.40, or 740 cents together, so we can make the algebraic equation:

5N + 25Q = 740

That's our first equation.  Now we double the number of nickels to 2N, and we increase the number of quarters to Q + 8.  2N nickels are worth 5(2N) = 10N cents, and Q + 8 quarters are worth 25(Q + 8) cents.  Together, this new set of nickels and quarters are worth $10.80, or 1080 cents, giving us a second algebraic equation:

10N + 25(Q + 8) = 1080
10N + 25Q + 200 = 1080 (distributing 25 on the left side)
10N + 25Q = 880 (subtracting 200 from both sides)

Now we have two linear equation in two unknowns.  We can solve this by a number of methods, but we'll try substitution.  First, notice that we can simplify the first equation by dividing both sides by 5:

N + 5Q = 148

Now we can solve for N:

N = 148 - 5Q (subtract 5Q from both sides)

Now we'll substitute 148 - 5Q for N in the second equation and solve for Q:

10(148 - 5Q) + 25Q = 880 (substituting)
1480 - 50Q + 25Q = 880 (distributing 10 on the left side)
1480 - 25Q = 880 (combining like terms)
-25Q = -600 (subtracting 1480 from both sides)
Q = 24 (dividing both sides by -25)

So Q = 24.  We can now substitute back to find N:

N = 148 - 5(24) = 148 - 120 = 28

So our solution is that Joe originally had 28 nickels and 24 quarters.  

Let's see if this makes sense.  28 nickels would be worth 140 cents, or $1.40, and 24 quarters would be worth 600 cents, or $6.00, making $1.40 + $6.00 = $7.40, which is what we want.  Now we double the number of nickels to 56 and increase the number of quarters by 8 to get 32.  56 nickels are worth 280 cents, or $2.80, and 32 quarters are worth 800 cents, or $8.00, making $2.80 + $8.00 = $10.80, which is also what we want.  So we have the correct answer.