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"Number" Word Problems

• The sum of two consecutive integers is 15. Find the numbers.

They've given you two pieces of information here. First, you know that you are adding two numbers, and that they total to fifteen. Secondly, the numbers are nice neat round numbers (like –3 or 6), not messy ones (like –4.628 or ¹⁷/₃₂), and that the second number is one more than the first. This last piece of information comes from the fact that "consecutive integers" (or "consecutive whole numbers", if they're restricting the possibilities to only positive numbers) are one apart. Examples of "consecutive integers" would be –12 and –11, 1 and 2, and 99 and 100. Using these facts, we can set up the translation. Let the first number be "n". Then the second number has to be "n + 1". The sum is then:

n + (n + 1) = 15
2n + 1 = 15
2n = 14
n = 7   Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved

Note that the problem did not ask for the value of the variable n; it asked for the identity of two numbers. So the answer is not "n = 7"; the actual answer is "The numbers are 7 and 8."

• The product of two consecutive negative even integers is 24. Find the numbers.

They have told us quite a bit about these two numbers. The numbers are even and negative. (The fact that they are negative may help if we come up with two solutions -- a positive and a negative -- so we'll know which one to pick.) Since even numbers are two apart (for example, –4 and –2 or 10 and 12), then we also know that the second number is two greater than the first. We also know that, when we multiply the two numbers, we get 24. In other words, letting the first number be "n" and the second number be "n + 2", we have:

(n)(n + 2) = 24
n² + 2n = 24
n² + 2n – 24 = 0
(n + 6)(n – 4) = 0

Then the solutions are n = –6 and n = 4. Since the numbers we are looking for are negative, we can ignore the "4" and take n = –6. Then n + 2 = –4. The answer is

The numbers are –6 and –4.

Do not assume that you can use both solutions if you just change signs to be whatever you feel like. While this often "works", it does not always work, and it's sure to annoy your teacher. Throw out invalid results, and solve properly for valid ones.

• Twice the larger of two numbers is three more than five times the smaller, and the sum of four times the larger and three times the smaller is 71. What are the numbers?

Before you get upset, remember that the point here is to give you practice in unwrapping and unwinding these words, and turning these words into algebraic equations. The point is in the solving, not in the relative "reality" of the problem. That said, how do you solve this? The best first step is to start labelling:

the larger number:  x
the smaller number:  y

twice the larger:  2x
three more than five times the smaller:  5y + 3
relationship between ("is"):  2x = 5y + 3

four times the larger:  4x
three times the smaller:  3y
relationship between ("sum of"):  4x + 3y = 71

Now you have two equations:

2x = 5y + 3
4x + 3y = 71

Solve, say, the first equation for x:

x = (5/2)y + (3/2)

Plug this into the second equation for "x":

4[ (5/2)y + (3/2) ] + 3y = 71
10y + 6 + 3y = 71
13y + 6 = 71
13y = 65
y = 65/13 = 5

Then you can solve for x:

x = (5/2)y + (3/2)
x = (5/2)(5) + (3/2)
x = (25/2) + (3/2)
x = 28/2 = 14

As always, remember to answer the question that was actually asked. The solution here is not "x = 14", but is the following sentence:

The larger number is 14, and the smaller number is 5.

The trick to doing this type of problem is to label everything very explicitly. Until you become used to doing these, do not attempt to keep track of things in your head. Do as I did in this last example: clearly label every single step. When you do this, these problems generally work out rather easily.

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