Find three consecutive integers whose sum is 132 and show work
Let's call the first of these consecutive integers "x."
Then we know that the second consecutive integer is "x+1" and the third consecutive integer is "x+2."
We also know that the sum of these three integers is 132:
x + (x+1) + (x+2) = 132
Now we can simplify and solve for x:
3x + 3 = 132
3x = 129
x = 43
So the first of the three consecutive integers is 43, and the next two are:
43+1 = 44
43+2 = 45
Therefore, the three consecutive integers whose sum is 132 are 43, 44, and 45.
Let's assume the first integer is x.
The next two consecutive integers would be x + 1 and x + 2.
The sum of these three integers can be expressed as:
x + (x + 1) + (x + 2) = 132
Simplifying the equation:
3x + 3 = 132
Subtracting 3 from both sides:
3x = 129
Dividing both sides by 3:
x = 43
So, the first integer is 43.
The next two consecutive integers are:
43 + 1 = 44
43 + 2 = 45
Therefore, the three consecutive integers whose sum is 132 are 43, 44, and 45.
To find three consecutive integers whose sum is 132, we can use algebra.
Let's assume the first integer is x. Since the integers are consecutive, the second integer would be x + 1, and the third integer would be x + 2.
According to the problem, the sum of these three consecutive integers is 132. So we can set up an equation:
x + (x + 1) + (x + 2) = 132
Simplifying this equation, we have:
3x + 3 = 132
Subtracting 3 from both sides:
3x = 129
Dividing both sides by 3:
x = 43
Now we know that the first integer is 43.
To find the second integer, we substitute x back into the expression x + 1:
43 + 1 = 44
So, the second integer is 44.
To find the third integer, we substitute x back into the expression x + 2:
43 + 2 = 45
Therefore, the three consecutive integers whose sum is 132 are 43, 44, and 45.