Question 215773
What are three consecutive integers such that the sum of twice the smallest and three times the largest is 126?


Step 1.  Let n be an integer (smallest integer), n+1 and n+2 (largest integer) be the next two consecutive integers.


Step 2.  Let 2n be twice the smallest integer.


Step 3.  Let 3(n+2) be  three times the largest integer.


Step 4.  Then {{{2n+3(n+2)=126}}} since the sum of twice the smallest and three times the largest is 126.  


Step 5.  The following steps will solve the equation in Step 4.  


{{{2n+3(n+2)=126}}}


{{{2n+3n+6=126}}}


{{{5n+6=126}}}


Subtract 6 from both sides of the equation


{{{5n+6-6=126-6}}}


{{{5n=120}}}


Divide 5 to both sides of the equation


{{{5n/5=120/5}}}


{{{n=24}}}


With {{{n=24}}}, then {{{n+1=25}}} and {{{n+3=26}}}


Check sum...{{{2*24+3*26=126}}} which is a true statement.


Step 6.  ANSWER:  The numbers are 24, 25, and 26.


I hope the above steps were helpful. 


For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J