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
6x + 5 = 53 6x = 48 x = 8 8 + 45 = 53 8 and 9
Let the integers be n and n + 1.n + 3(n+ 1) = 47 4n + 3 = 47 4n = 44 n = 11 n+ 1 = 12.
Let the integers by 'm' & 'm+1' m +3(m+1) = 23 m + 3m + 3 = 23 4m = 20 m = 5 Hence m+1 = 6
Let x equal the smaller integer. Three times the larger is 3(x + 1) or 3x + 3. Added all together, 4x + 3 = 31. Solve for x. 4x = 28 x = 7 Check it. 7 + 24 = 31 It checks.
The sum of any three consecutive odd integers is going to give an odd result. It is impossible for the sum of an odd number of odd integers to equal an even number.
12 and 13.
The integers are 10 and 11.
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
There is not such set of integers.
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.
Let the integers be n and n + 1.n + 3(n+ 1) = 47 4n + 3 = 47 4n = 44 n = 11 n+ 1 = 12.
There is no set of two consecutive integers having a product of 14. Product means the result of multiplication.