Often, word problems appear confusing, and it is difficult to know where to begin. Here are some steps that will make solving word problems easier:

1. Read the problem.
2. Determine what is known and what needs to be found (what is unknown).
3. Try a few numbers to get a general idea of what the solution could be.
4. Write an equation.
5. Solve the equation by inverse operations or by plugging in values.
6. Check your solution--does it satisfy the equation? Does it make sense in the context of the problem? (e.g. A length should not be negative.)

Example 1: Matt has 12 nickels. All the rest of his coins are dimes. He has just enough money to buy 2 slices of pizza for 95 cents each. How many dimes does he have?

1. Read the problem.
2. What is known? Matt has 12(5) = 60 cents in nickels. Matt has 2(95) = 190 cents total. What needs to be found? The number of dimes that Matt has.
3. Try a few numbers:
   
   5 dimes? 10(5) + 60 = 110. Too low.
   10 dimes? 10(10) + 60 = 160. Still too low.
   20 dimes? 10(20) + 60 = 260. Too high.
   
   So we know the answer is between 10 and 20.
4. Write an equation: 10d + 60 = 190 where d is the number of dimes Matt has.
5. Solve using inverse operations:
   
   10d + 60 - 60 = 190 - 60
   10d = 130
   =
   d = 13
6. Check: 10(13) + 60 = 190? Yes. Does 13 dimes make sense in the context of the problem? Yes.

Thus, Matt has 13 dimes.

Example 2: Jen is shooting free-throws on the basketball court. She makes 85% of her shots. If she makes 51 shots, how many does she miss?

1. Read the problem.
2. What is known? Jen makes 85% -- or -- of her shots. Jen makes 51 shots. What needs to be found? The number of shots that Jen misses.
3. Try a few numbers:
   
   5 shots? = . Not enough misses.
   10 shots? = . Too many misses.
   So we know the answer is between 5 and 10.
4. Write an equation: = where x is the number of misses.
5. Solve using inverse operations:
   
   =
   51() = 51()
   51 + x = 60
   51 + x - 51 = 60 - 51
   x = 9
6. Check: = ? Yes. Does 9 shots make sense in the context of the problem? Yes.

Thus, Jen misses 9 shots.

Example 3: The area of this square is 2 times its perimeter. How long is a side?

1. Read the problem.
2. What is known? The area of the square is 2 times its perimeter. The formula for area is A = x² and the formula for perimeter is p = 4x. What needs to be found? The length of a side.
3. Try a few numbers:
   x = 5? A = 5² = 25, p = 4(5) = 20. Area too small.
   x = 10? A = 10² = 100, p = 4(10) = 40. Area too large.
   
   So we know the answer is between 5 and 10.
4. Write an equation: x² = 2(4x). x² = 8x
5. Solve by plugging in values----or by using inverse operations:
   
   =
   x = 8
6. Check: 8² = 8(8) ? Yes. Does 8 make sense in the context of the problem? Yes.

Thus, the square has side length 8.