Find two consecutive integers such that the larger is nine more than twice the smaller

Answer 1

Let n be "an integer" then the next consecutive integer is (n + 1)

So twice the smaller is 2n and 9 more than that is 2n + 9 so:

2n + 9 = n + 1

n + 9 = 1

n = -8

The next consecutive number is -7