Two consecutive two-digit numbers are added. The digits of this sum, written in the reverse order, represent the largest of the consecutive numbers. What is the smaller number ?

The smaller of the consecutive number is 36

Step-by-step explanation:

Let the digits be ab, and ab + 1, therefore, we are given;

The given parameters are;

 ab

+ ab + 1

b+1   a

Therefore, we note that adding the two unit numbers b and b + 1, will give a number larger than 10, from which we carry the number 1 on the tens side to be added to the  tens unit of when adding the tens unit side of the two numbers as follows;

b + b + 1 = 10 + a...(1)

1 + a + a = b + 1...(2)

From equation (2), we have;

1 + a + a = b + 1

b + 1 = 1 + a + a

b = 2·a...(3)

Substituting the value of b from equation (3) in equation (1) gives;

b + b + 1 = 10 + a

2·a + 2·a + 1 = 10 + a

4·a - a = 10 - 1

3·a = 9

a = 9/3 = 3

a = 3

From equation (3) b = 2·a = 2 × 3 = 6

b = 6

The numbers are a b, and a b + 1 which are 36 and 36 + 1 = 37

We check to get;

36 + 37 = 73

Arranging 73 in reverse order = 37 which is the largest of the consecutive numbers

Therefore, the smaller of the consecutive number = 36.

the correct answer is 45x

9×5=45 then add the x

160

step-by-step explanation:

idk

i hope i