There are no such integers.
The integers are 10 and 11.
55 and 57
x+5(x+1)=53 6x+5=53 6x=48 x=8 numbers=8 and 9
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.
There are no such integers.
12 and 13.
The integers are 10 and 11.
10,12
They are 5 and 6 because 5+(2*6) = 17
Let the smaller be n, then the larger is n+1; and: n + 4(n+1) = 59 → n + 4n + 4 = 59 → 5n = 55 → n = 11 → the two consecutive integers are 11 and 12.
6x + 5 = 53 6x = 48 x = 8 8 + 45 = 53 8 and 9
55 and 57
12 and 13.
x+5(x+1)=53 6x+5=53 6x=48 x=8 numbers=8 and 9
this is a algebraic word problem: we know we have two consecutive integers so x + 1 = y and we know that the smaller one (x) added to three times the larger, the results is 23 so x + 3y = 23 substitute (x+1) for y: x + 3(x+1) = 23 4x + 3 = 23 4x = 20 x = 5 y = 6
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.