and since we know now what "x", let's use that in either equation, say the first equation,
another way to do it is by "elimination", depending on the equations, one way may work quicker than the other, in this case, substitution was the simpler one, because the first equation is already solved for "y", so we just grabbed it and plugged it in the 2nd equation.
Step-by-step explanation:
We are picking 6 numbers from the numbers 1,2,3,4,5,6,7,8,9,10. Since we care about numbers being next to each other, we might think of the 10 numbers as being distributed in 5 boxes (which you can think of as the holes):
| 1 2 | 3 4 | 5 6 | 7 8 | 9 10 |
So on the first box we have the numbers 1 and 2, on the second box we have the numbers 3 and 4, and so on. Since we are picking 6 numbers from those 10 numbers, that means we'll have to pick 6 boxes (and inside each box we pick a number), but we only have 5 available boxes, so by the pigeonhole principle, we'll have to pick 1 same box at least two times. Since on each picked box we'll need to pick a number, on this box which was picked two times, we will have to pick both of its numbers. And so those 2 numbers inside that box will be next to each other (meaning they're consecutive numbers).
I think it’s A because 4*2 is 8 and if n were quadrupled it would be 2
