SOLUTION: four consecutive integers whose sum is 58

Question 218078: four consecutive integers whose sum is 58
Found 2 solutions by joma, drj:
Answer by joma(3) (Show Source): You can put this solution on YOUR website!
let x= 1st integer
let x+1= 2nd integer
x+2 be the 3rd integer
x+3 be the 4th "
the equation for this will be:
x+x+1+x+2+x+3= 58
so.. (combine like terms)
4x+6=58
4x=58-6
4x=52
x=13 (1st integer)
x+1=14
x+2=15
x+3=16
13+14+15+16=58

.. jOms..
(is this correct??)
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
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.

Divide 4 to both sides of the equation.

Then , and

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

http://www.FreedomUniversity.TV

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