• Jan 8th 2009, 09:36 PM
anabanana
Missed today's class due to being stuck sick in bed. I have a quiz tomorrow on Consecutive Integer Problems using fractions in the word problems. Can you help me with the following question? The larger of two consecutive odd integers is 4 less than 1/3 the smaller. Find the numbers.
Thanks(Nerd)
• Jan 8th 2009, 10:45 PM
Constatine11
Quote:

Originally Posted by anabanana
Missed today's class due to being stuck sick in bed. I have a quiz tomorrow on Consecutive Integer Problems using fractions in the word problems. Can you help me with the following question? The larger of two consecutive odd integers is 4 less than 1/3 the smaller. Find the numbers.
Thanks(Nerd)

Two consecutive odd integers always take the form $\displaystyle 2k+1,\ 2k+3$ for some integer $\displaystyle k$.

You are told that

$\displaystyle 2k+3=\frac{(2k+1)}{3} - 4$

Which you now solve for $\displaystyle k$, and hence for the two numbers (don't be supprised if one or both are negative)

.
• Jan 8th 2009, 10:49 PM
Integers
Hello anabanana
Quote:

Originally Posted by anabanana
Missed today's class due to being stuck sick in bed. I have a quiz tomorrow on Consecutive Integer Problems using fractions in the word problems. Can you help me with the following question? The larger of two consecutive odd integers is 4 less than 1/3 the smaller. Find the numbers.
Thanks(Nerd)

Suppose that the smaller odd integer is $\displaystyle n$; then the next larger one will be $\displaystyle (n+2)$.

So: $\displaystyle (n+2)$ is $\displaystyle 4$ less than $\displaystyle \frac{1}{3}$ of $\displaystyle n$.

In other words: $\displaystyle n+2 = \frac{1}{3}n - 4$

Add $\displaystyle 4$ to both sides; multiply both sides by $\displaystyle 3$; then ...?

Can you solve it now? (The answer is a negative odd integer.)

• Jan 9th 2009, 08:54 AM
anabanana
Problem solved
Thank you! I should ace this quiz today.(Happy)