Question 216388
Find three consecutive odd integers such that three times the second minus the third is 13 more than the first. 


Step 1.  Let {{{n}}} be one odd integer, let {{{n+2}}} be second consecutive integer and let {{{n+4}}} be the third consecutive odd integers.


Step 2.  Let {{{3(n+2)}}}  three times the second.


Step 3.  Then {{{3(n+2)-(n+4)=n+13}}} since three times the second minus the third is 13 more than the first.


Step 4.  The following steps will solve the equation in Step 3.


{{{3(n+2)-(n+4)=n+13}}}


{{{3n+6-n-4=n+13}}}


{{{2n+2=n+13}}}


Subtract n+2 to both sides of the equation


{{{2n+2-n-2=n+13-2}}}


{{{n=11}}}


Then {{{n+2=13}}}  and {{{n+4=15}}}


Check...{{{3(n+2)-(n+4)=n+13}}} then {{{3*13-15=11+13}}} or {{{24=24}}} which is a true statement.


Step 5.  The three consecutive odd integers are 11, 13, and 15.


I hope the above steps and explanation were helpful. 


For Step-By-Step videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry please visit http://www.FreedomUniversity.TV/courses/Trignometry. 


Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs.


Respectfully, 
Dr J