Question

Find two consecutive integers such that greater of the two is 22 more the twice than smaller

Answer 1

Let the first integer = a

The next integer will be (a + 1)

Consecutive integers differ by 1

Now, since the greater of the two, (a + 1), is 22 more than twice the smaller, we can write an equation for this statement, and the equation is;

(a + 1) - 22 = 2a

a - 2a = 22 - 1

-a = 21

a = -21

and the next integer;

a + 1 = -21 + 1 = -20

So, the integers are -21 and -20

Answer 2

Solution :-

Let the smaller integer be x 

Then, the greater integer will be (x + 1)

The greater integer (x + 1) is 22 more than twice the smaller number (x).

Now, according to the question.

⇒ (x + 1) = 2x + 22

⇒ x - 2x = 22 - 1

⇒ - x = 21

⇒ x = - 21

So, smaller integer is - 21

Greater integer = x + 1

⇒ - 21 + 1

= - 20 

So, the greater integer is - 20

Hence, smaller integer is - 21 and greater integer is - 20
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Now, let us check our answer.

x + 1 = 2x + 15

Substituting for x + 1 and x

⇒ - 21 + 1 = 2*(- 21) + 22

⇒ - 20 = - 42 + 22

⇒ - 20 = - 20

L.H.S. = R.H.S. 

So our answer is correct.