You can put this solution on YOUR website! Question:
Three consecutive integers are such that the sum of the first and third is 11 less than three times the second. Find the numbers.
Answer:
Here you have three consecutive number...
can find out the th
So you can take them as, a, a+1 and a+2
Sum of the first and the third is a + ( a+2)
Three times the second can be written as 3(a+1)
It is given that, the sum of the first and third is 11 less than three times the second
So... a + ( a+2) = 3(a+1)-11
==> a + a + 2= 3a + 3 -11
==> 2a + 2 = 3a + 3 -11
Subtract 3a from both sides..
==> 2a +2 - 3a = 3a -8 - 3a
==> -1a + 2 = -8
==> -1a + 2 - 2 = -8 - 2
==> -1a = -10
Divide both sides by -1
==>
==>
a = 10
a+1 = 10 +1 = 11
a + 2 = 10 + 2 = 12
So 10,11 and 12 are the three consecutive numbers.
