Ten Times the smallest of three Consecutive Intgers Is Twenty Two More Than Three Times the sum of the integers .Find The integers?
Twety times the smallest of three consecutive integers is sixty eight more than four time the sum of the integers. Find the number
If the three numbers are x, x+1 and x+2, then
10x = 22+3(x+x+1+x+2)
Now solve for x.
Now, Why All The Capital Letters?
To solve this problem, let's break it down step by step:
Let's assume the three consecutive integers as x, (x+1), and (x+2).
According to the given information:
Ten times the smallest of the three consecutive integers is twenty-two more than three times the sum of the integers:
10x = 3(x + (x+1) + (x+2)) + 22
Now let's simplify the equation:
10x = 3(3x + 3) + 22
10x = 9x + 9 + 22
10x = 9x + 31
Next, we want to isolate the variable x by moving the 9x term to the other side:
10x - 9x = 31
x = 31
Therefore, the first integer is 31.
Since we know that the consecutive integers are x, (x+1), and (x+2), we can find the other two integers:
First integer: 31
Second integer: 31 + 1 = 32
Third integer: 31 + 2 = 33
So the three consecutive integers are 31, 32, and 33.