Answer:
The integers are 11, 12, 13
Step-by-step explanation:
Given:
Five times the smallest of three consecutive integers is 17 less than twice the sum of the integers
To Find:
The integers = ?
Solution:
Let the 3 consecutive integer be
n , n+1 , n+2
then
5 times the smallest of the integer. i
here the smallest integer is n
5n = 17 less than twice the sum of the integers
The sum of the integer is
=> n+n+1+n+2
Now
5n = 17 less than twice( n+n+1+n+2)
5n = 17 less than 2( n+n+1+n+2)
5n = 2( n+n+1+n+2) - 17
Solving the equation,
5n = 2( 3n +3) - 17
5n = 6n + 6 - 17
5n -6n = 6 - 17
-n = -11
n =11
Then
n+1 = 11+1 = 12
n+2 = 11+2 =13
The integers are 11,12 and 13
Step-by-step explanation:
Let x,x+1,x+2 be the three consecutive integers. Let x be the smallest of three consecutive integers.
Then the expression can be written as
[tex]5x=2(x+x+1+x+2)-17[/tex]
Adding the x terms, we get,
[tex]5x=2(3x+3)-17[/tex]
Multiply the term [tex]3x+3[/tex] by 2,
[tex]5x=6x+6-17[/tex]
Adding the constant terms, we get,
[tex]5x=6x-11[/tex]
Subtracting 6x from both sides of the equation, we get,
[tex]-x=-11[/tex]
Thus, [tex]x=11[/tex]
Substituting the value of x in x+1 and x+2, we get the value of three consecutive integers.
[tex]x+1=11+1=12\\x+2=11+2=13[/tex]
Thus, the three consecutive integers are [tex]11,12 and 13[/tex]
