the sum of 2 consecutive intergers is one less than twice the greater integer?

Question

translate into a variable expression or equation

Answer 1

(x+(x+1)) = 2(x+1)-1

Answer 2

n + (n+1) = 2 (n+1) - 1

Answer 3

x + (x+1) = 2(x+1) - 1

Answer 4

x + (x+1) = 2(x+1) -1

Answer 5

Let a be the first integer. a + 1 is the second integer. So:

a + a + 1 + 1 = 2(a + 1)
2a + 2 = 2a + 2

Note that this reduces to 0=0. That means any integer value for a will satisfy this equation.

Answer 6

If n is the smaller integer, the greater is n +1. Then, in equation, we have

n + n +1 = 2(n +1) -1. This is equivalent to

2n +1 = 2n +2 -1= 2n +1, so we have an identity, this condition holds for every integer n.

Answer 7

Let x, x+1 be the consecutive integers.
sum= x+(x+1)=2(x+1)-1
2x+1=2x+2-1
2x+1=2x+1

Answer 8

x + (x+1) = 2(x + 1) - 1
2x + 1 = 2x + 1,
true for all integers.

Answer 9

(x -1) + x = 2x -1