# The sum of the digits of a certain two-digit number is 11. Reversing its digits decreases the...

## Question:

The sum of the digits of a certain two-digit number is 11. Reversing its digits decreases the number by 63.

What is the system?

## Elimination Method:

A system of two equations can be solved in many methods. One of the methods is the "elimination method". In this method, we add or subtract the equations to get a variable canceled. We will solve the resultant equation 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 decreases by {eq}63 {/eq}".

So, we get:

$$\text{Reversed number}= \text{Original number} -63 \\ 10y+x = (10x+y)-63 \\ \text{Subtracting 10x and y from both sides}, \\ -9x+9y =-63 \\ \text{Dividing both sides by 9},\\ -x+y=-7 \,\,\,\,\,\,\,\rightarrow (1)$$

The problem also says, "the sum of its digits is {eq}11 {/eq}".

So, we get:

$$x+y=11 \,\,\,\,\,\,\,\rightarrow (2)$$

$$2y=4 \\ \text{Dividing both sides by 2}, \\ y=2$$

Substitute this in (2):

$$x+2=11\\ x=9$$

Therefore, the system is:

{eq}\boxed{\mathbf{-x+y=-7\\ x+y=11}} {/eq}

and the required number is {eq}xy = \boxed{\mathbf{92}} {/eq}.