Step 1. Let n be an integer.
Step 2. Let n+1 and n+2 be the next two consecutive integers.
Step 3. Let n+n+1=2n+1 be the sum of the two smallest integers.
Step 4. Let 2(2n+1) be twice the sum of the two smallest integers.
Step 5. Let 3(n+2) be three times the largest number
Step 6. Then using Steps 3 and 4, 2(2n+1)=3(n+2)+10 since twice the sum of the two smallest numbers is 10 more than three times the largest number.
Step 7. Solving yields the following steps:
| Solved by pluggable solver: EXPLAIN simplification of an expression | |
Your Result: YOUR ANSWERDETAILED EXPLANATIONLook at Moved these terms to the left It becomes Look at Expanded term It becomes Look at Multiplied numerator integers It becomes Look at Multiplied numerator integers It becomes Look at Added fractions or integers together It becomes Look at Moved It becomes Look at Removed extra sign in front of It becomes Look at Expanded term It becomes Look at Multiplied numerator integers It becomes Look at Added fractions or integers together It becomes Look at Removed extra sign in front of It becomes Look at Eliminated similar terms It becomes Look at Remove extraneous '1' from product It becomes Look at Solved linear equation It becomes Result: This is an equation! Solutions: n=14. Universal Simplifier and SolverDone! |
n=14, then n+1=15 and n+2=16
Check original equation
Step 8. ANSWER: The three consecutive integers are 14, 15, and 16.
I hope the above steps were helpful.
For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
Good luck in your studies!
Respectfully,
Dr J
I hope the above steps were helpful.
For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
Good luck in your studies!
Respectfully,
Dr J