Question 212405
Find two consecutive odd integers such that 3 times the first is 5 more than twice the second




Step 1.  Let n be one odd integer.


Step 2.  Let n+2 be the next odd integer.


Step 3.  Generate an equation for the given problem statement: Find two consecutive odd integers such that 3 times the first IS 5 more than twice the second.  3n is one side of equation and 5+2(n+2) is the opposite side of equation.  "is" in the given problem statement implies "equal to" so


{{{3n = 5+2(n+2)}}}


{{{3n = 5+2n+4}}}


Step 4.  Solve for n.  Subtract 2n from both sides of equation.


{{{3n-2n = 5+2n+4-2n}}}


{{{n = 9}}}


{{{n+2 = 11}}}


Check numbers with problem statement. 


{{{3(n)=3(9)=27}}}


{{{2(n+2)=2(11)=22}}} so 27 is 5 more than twice 11.  So numbers are odd and satisfies problem statement.


Step 5.  ANSWER:  Numbers are 9 and 11.



Hope the steps were helpful.  Good luck in your studies.   
Respectfully, 
Dr J


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