# The sum of a certain two-digit number is 7. Reversing its digits increases the number by 27. What...

## Question:

The sum of a certain two-digit number is 7. Reversing its digits increases the number by 27. What is the number?

## Elimination Method:

The elimination method is one of the methods of solving a system of two equations. In this method, we add or subtract the equations so as to get one variable canceled. Then we get a linear equation in one variable which can easily be solved. We can then substitute that value 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}27 {/eq}".

So, we get:

\begin{align} \text{Reversed number}&= \text{Original number}+27 \\[0.2cm] 10y+x &= (10x+y)+27 \\ \end{align}

Subtracting {eq}10x{/eq} and {eq}y{/eq} from both sides,

$$-9x+9y =27$$

Dividing both sides by {eq}9{/eq},

$$-x+y=3 \,\,\,\,\,\,\,\rightarrow (1)$$

The problem also says, "the sum of its digits is {eq}7 {/eq}".

So, we get:

$$x+y=7 \,\,\,\,\,\,\,\rightarrow (2)$$

$$2y=10$$

Dividing both sides by {eq}2{/eq},

$$y=5$$

Substitute this in {eq}(2){/eq}:

$$x+5=7\\ x=2$$

Therefore, the required number is {eq}xy = \color{blue}{\boxed{\mathbf{25}}} {/eq}.