x = smallest
x+2 = largest
x +4(x+2) = 28
5x +8 = 28
5x = 20
x =4 (smaller)
X+2 = 6 (larger)
12 and 13.
58 = 3x + yy = x-258 = 3x + x-258 = 4x - 256 = 4x14 = xx-2 = yy = 16Let x= the lesser integer=14Let x+2 = the larger integer=163(x)+(x+2)=583x+x+2=584x+2=584x=56x=14x+2(14)+216Let the smaller number be n, then the larger number is n + 2.3n + (n + 2) = 58 : 4n + 2 = 58 : 4n = 56 : n = 14. then n + 2 = 16The two consecutive even numbers are 14 and 16.
88
When equals are added to equals, you don't necessarily get wholes. But whatever you do get, they're equal.
Any irrational number, when added to 13, will produce an irrational number.
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.