# How do you find two consecutive odd integers such that 4 times the larger is 29 more than 3 times the larger?

Mar 1, 2016

Page numbers are 27 and 29

#### Explanation:

$\textcolor{b l u e}{\text{Build and expression that guarantees the numbers are odd}}$

Let any number be $n$

Then $2 n$ will always be even. So $2 n + 1$ will always be odd
The next odd number will be $\left(2 n + 1\right) + 2 = 2 n + 3$

So our two consecutive numbers can be represented by

$2 n + 1 \text{ and } 2 n + 3$

$\textcolor{g r e e n}{\text{Note that the value of n is what I call a seed value.}}$
color(green)("It is not a page number"!)
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Building the equation of relationship}}$

Breaking down the question into its parts

"4 times the larger" $\text{ } \to \textcolor{b r o w n}{4 \left(2 n + 3\right)}$
"is "$\text{ } \to \textcolor{b r o w n}{4 \left(2 n + 3\right) =}$
"29 more than "$\text{ } \to \textcolor{b r o w n}{4 \left(2 n + 3\right) = 29 +}$
"3 times the larger "$\text{ } \to \textcolor{b r o w n}{4 \left(2 n + 3\right) = 29 + 3 \left(2 n + 3\right)}$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply out the brackets giving:

$8 n + 12 = 29 + 6 n + 9$

Collecting like terms

$8 n - 6 n = 29 + 9 - 12$

$2 n = 26$

$n = 13 \textcolor{g r e e n}{\text{ This is the seed value}}$

color(blue)("The first page is " 2n+1 -> 2(13)+1 = 27

color(blue)("The second page is 2 pages on " -> 27+2=29