Four times the sum of two consecutive integers equals 2 more than 14 times the larger integer?

Question

Find the integers.

Answer 1

heres how I read it::


4(x+(x+1)) = 14(x+1) + 2


and now its just solving::

4(x + (x+1) ) = 14(x+1) + 2

4(2x + 1) = 14x + 14 + 2

8x + 4 = 14x + 16

-6x = 12

x = -2

that's the first one.
the second one, the next consecutive integer, the x + 1, is literally just x + 1

-2 + 1 = -1


-2 and -1

Answer 2

let the consecutive integers be x and x+1.

4( x+ x+ 1) = 2 + 14(x+1)

4( 2x + 1) = 2+ 14x +14

8x + 4 = 16 + 14x

8x - 14x = 16 - 4

-6x = 12

x = -2.

x+1 = -2 +1 = -1.

Answer 3

What you wrote can be expressed as

4 ( n + n + 1) = 14 (n + 1) + 2

8n + 4 = 14n + 16

8n - 14n = 16 - 4

-6n = 12

n = 12 / -6 = -2

If n = -2, n + 1 = -1

Let us see whether the conditions are fulfilled.

-2 + (-1) = -2 - 1 = - 3

4 times - 3 is - 12

The larger integer is -1 (though it may be confusing)

and 14 times that is -14 and 2 more than that is -12.

So, our calculation is correct. -2 and -1 are the two integers.