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

`color(blue)("Build and expression that guarantees the numbers are odd")`

Let any number be `n`

Then `2n` will always be even. So `2n+1` will always be odd
The next odd number will be `(2n+1)+2 = 2n+3`

So our two consecutive numbers can be represented by

`2n+1" and "2n+3`

`color(green)("Note that the value of n is what I call a seed value.")`
`color(green)("It is not a page number"!)`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Building the equation of relationship")`

Breaking down the question into its parts

"4 times the larger" `" "->color(brown)(4(2n+3))`
"is "`" " ->color(brown)(4(2n+3)=)`
"29 more than "`" " ->color(brown)(4(2n+3)=29 +)`
"3 times the larger "`" "->color(brown)(4(2n+3)=29 +3(2n+3))`
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Multiply out the brackets giving:

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

Collecting like terms

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

`2n=26`

`n=13 color(green)(" 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`