Let the integers by 'm' & 'm+1'
m +3(m+1) = 23
m + 3m + 3 = 23
4m = 20
m = 5
Hence m+1 = 6
There are no two consecutive even integers, consecutive odd integers, or consecutive integers that satisfy that relationship.
The smaller integer is 6, the larger integer is 32
Let the smaller integer be x, then then larger integer is x + 2, and: 3x + (x + 2) = 58 → 4x = 56 → x = 14 → The two integers are 14 and 16.
44 & 45
Let x = 1st integer, since consecutive even integer differs by 2 then the 2nd integer = x + 2. So we have, x + 4(x + 2) = 48 x + 4x + 8 = 48 5x = 40 x = 8 (1st integer) Thus, the integers are 8 and 10. Check.
The larger integer is 30. The smaller is 28.
-1
x+3 and x+4 would be consecutive integers.
There can be no such integers: a smaller integer cannot be 5 times the larger number.
The product of two consecutive positive integers can be found by multiplying the smaller integer by the larger integer. If the smaller integer is represented as ( n ), then the larger integer would be ( n + 1 ). Therefore, the product of two consecutive positive integers is ( n \times (n + 1) ).
the sum of two consecutive integers is -241, what is the larger integer?
5
114.As we know that the numbers are 2 consecutive even integers, we know that one number will be 2 larger than the other. We can use this to solve the problem with algebra:Where x is the smaller number,x + (x + 2) = 2262x + 2 = 2262x = 224x = 112Therefore the larger integer is 112 + 2 = 114
There are no two consecutive even integers, consecutive odd integers, or consecutive integers that satisfy that relationship.
There are no such integers.
3*(x+1) - x = 7 So 3x + 3 - x = 7 then 2x = 4 or x = 2 So the two integers are 2 and 3
Suppose the smaller integer is x, then the larger one is x+1. x + 3*(x+1) = 43 That is x + 3x + 3 = 43 so that 4x = 40 and that implies that x = 10 and so the other integer is 11.