Strategies for Solving Word Problems with Variables

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:

Read the problem.

Determine what is known and what needs to be found (what is unknown).

Try a few numbers to get a general idea of what the solution could be.

Write an equation.

Solve the equation by inverse operations or by plugging in values.

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?

Read the problem.

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.

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.

Write an equation: 10d + 60 = 190 where d is the number of dimes Matt has.

Solve using inverse operations:

10d + 60 - 60 = 190 - 60 10d = 130 = d = 13

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?

Read the problem.

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.

Try a few numbers:

5 shots? = . Not enough misses. 10 shots? = . Too many misses. So we know the answer is between 5 and 10.

Write an equation: = where x is the number of misses.

Solve using inverse operations:

= 51() = 51() 51 + x = 60 51 + x - 51 = 60 - 51 x = 9

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?

Read the problem.

What is known? The area of the square is 2 times its perimeter. The formula for area is A = x^{2} and the formula for perimeter is p = 4x. What needs to be found? The length of a side.

Try a few numbers:

x = 5? A = 5^{2} = 25, p = 4(5) = 20. Area too small. x = 10? A = 10^{2} = 100, p = 4(10) = 40. Area too large.

So we know the answer is between 5 and 10.

Write an equation: x^{2} = 2(4x). x^{2} = 8x

Solve by plugging in values----or by using inverse operations:

= x = 8

Check: 8^{2} = 8(8) ? Yes. Does 8 make sense in the context of the problem? Yes.