# The sum of the digits of a two-digit number is 8. When the digits are reversed, the number...

## Question:

The sum of the digits of a two-digit number is 8. When the digits are reversed, the number increases by 36.

Find the original number.

## Elimination Method:

(i) The elimination method is used to solve a system of two equations in two unknowns (variables).

(ii) In this method, we add or subtract the given equations and eliminate one variable.

(iii) We solve the resultant equation in one variable using the algebraic operations.

(iv) We can then substitute the value of this known variable in one of the given two equations and solve for the other variable.

Let us assume the required two-digit number to be {eq}xy {/eq}.

Here, {eq}x {/eq} is in the ten's place and {eq}y {/eq} is in the one's place.

So, the two-digit number = {eq}10x+y {/eq}.

The two-digit number when the digits are reversed is {eq}yx {/eq}.

In the same way explained above, the reversed number = {eq}10y+x {/eq}.

The problem says, "When the digits are reversed, the number increases by {eq}36 {/eq}".

So, we get:

$$\text{Reversed number}= \text{Original number}+36 \\ 10y+x = (10x+y)+36 \\ \text{Subtracting 10x and y from both sides}, \\ -9x+9y =36\\ \text{Dividing both sides by 9},\\ -x+y=4 \,\,\,\,\,\,\,\rightarrow (1)$$

The problem also says, "the sum of its digits is {eq}8 {/eq}".

So, we get:

$$x+y=8 \,\,\,\,\,\,\,\rightarrow (2)$$

$$2y=12 \\ \text{Dividing both sides by 2}, \\ y=6$$

Substitute this in (2):

$$x+6=8\\ x=2$$

Therefore, the required number is {eq}xy = \boxed{\mathbf{26}} {/eq}.