A.Write and solve an equation to find three consecutive integers with a sum of 126. Let n= the first integer.
B. In part A, could you solve the problem by letting n= the middle integer,n-1= the smallest integer, and n+1= the largest integer?
n+n+1+ n+2=126
3n=126-3
n=41, next 42,next 43
yes, you can do it that way.
N-1+n+n+1=`26
3n=126
n=42
so numbers are 41,42,43
A. To find three consecutive integers with a sum of 126, we can let n be the first integer. The second consecutive integer would then be n+1, and the third consecutive integer would be n+2. The sum of these three integers can be represented as:
n + (n+1) + (n+2) = 126
Simplifying the equation, we have:
3n + 3 = 126
Subtracting 3 from both sides:
3n = 123
Dividing both sides by 3:
n = 41
So, the three consecutive integers with a sum of 126 are 41, 42, and 43.
B. Yes, it is possible to solve the problem by letting n be the middle integer, n-1 be the smallest integer, and n+1 be the largest integer. However, it would require a slightly different approach. We can set up the equation as follows:
(n-1) + n + (n+1) = 126
Simplifying the equation, we have:
3n = 126
Dividing both sides by 3:
n = 42
So, when n is the middle integer, the three consecutive integers with a sum of 126 are 41, 42, and 43.