Question

HELP ASAP

Answer 1

Answer:

numbers are 13 and 14

Step-by-step explanation:

Let the numbers be y and y+1

The products of the numbers is written as:

y(y + 1)

The sum of the numbers is written as:

y + (y + 1) = 2y + 1

From the question,

The products of the two number is great than the sum by 155. This can be written as:

y(y + 1) = 2y + 1 + 155

y^2 + y = 2y + 156

y^2 + y — 2y —156 = 0

y^2 — y —156 = 0

Multiply the first term (i.e y^2) and the last term (—156) together. This gives —156y^2.

Now find two factors of —156y^2 such that when you add the two factors together, it will result to the 2nd term (ie —y) in the equation. These numbers are 12y and — 13y

Now substitute these numbers (i.e 12y and — 13y) in place of —y in the equation above

y^2 — y —156 = 0

y^2 + 12y — 13y —156 = 0

y(y + 12) — 13(y + 12) = 0

(y —13)(y + 12) = 0

y —13 =0 or y +12 =0

y = 13 or y = —12

Since the numbers are positive numbers, y =13

The first number = y = 13

The 2nd number = y + 1 = 13 +1 = 14

Therefore, the consecutive numbers are 13 and 14

Answer 2

Let s = the smaller one of the two.
Since they are consecutive, the second one is (s + 1)

Since they are natural numbers, they cannot be negative.

We know that s(s + 1) = s + (s + 1) + 155

So:
s(s + 1) = 2s + 156
s^2 + s = 2s + 156
s^2 - s - 156 = 0
(s - 13)(s + 12) = 0
From this, it looks like s can be either 13 or -12. But since it's a natural number, it cannot be negative. So, it is 13.

s was the smaller one. The larger one is s + 1 = 14
Check: 13*14=182=13+14+155