4+(6+6)
All these words in mathematics are represented by an equation. Write and solve it to find those integers.
If the first integer is x, the second one will be x + 2 (since consecutive even integers differ by 2). So we have:
x + 2(x +2) = 16
x + 2x + 4 = 16
3x + 4 = 16 subtract 4 from both sides
3x = 12 divide by 3 both sides
x = 4
Thus the integers are 4 and 6.
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.
Suppose n is the smaller integer. Then the larger integer is n + 1, so that 5 times the larger is 5*(n + 1). Their sum is n + 5*(n + 1) = 6n + 5 Therefore 6n + 5 = 41 6n = 41 - 5 = 36 So that n = 36/6 = 6
Let x represent the smaller of the two integers.Since integers are also members of the set of whole numbers, then the next largest consecutive integer must be (x+1)Translating the question into a mathematical equation:x + 2(x + 1) = 26x + 2x + 2 = 26add the "ex's"3x + 2 = 26subtract 2 from both sides of the equation3x = 24divide both sides of the equation by 3, to solve for xx = 8, and x + 1 = 9the solution is 8 and 9 [8 + (2*9)] = 26;; [8 + 18] = 26
Larger.
There are no such integers.
The integers are 10 and 11.
12 and 13.
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.