Question 218078
Four consecutive integers whose sum is 58.


Step 1.  Let n be an integer


Step 2.  Let n+1, n+2, and n+3 be the next three consecutive integers.


Step 3.  Then, n+n+1+n+2+n+3=58 since the sum of four consecutive integers is 58.


Step 4.  Solving, 4n+6=58, yields the following steps:


Subtract 6 from both sides of the equation.


{{{4n+6-6=58-6}}}


{{{4n=52}}}


Divide 4 to both sides of the equation.


{{{4n/4=52/4}}}


{{{n=13}}}  Then {{{n+1=14}}}, {{{n+2=15}}}  and  {{{n+3=16}}}


Check if sum is 58....13+14+15+16=29+29=58....which is a true statement.


Step 5.  ANSWER:  Numbers are 13, 14, 15, and 16.



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/Trigonometry. 


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


Respectfully, 
Dr J


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