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

1 Answer
Mar 1, 2016

Answer:

Page numbers are 27 and 29

Explanation:

#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#