# A two-digit number is 3 more than 4 times the sum of its digits.

Question:

A two-digit number is 3 more than 4 times the sum of its digits. If 8 is added to the number, the digits are reversed. Find the number.

Solution:

Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$.

The number is 3 more than 4 times the sum of the two digits. Thus, we have

$10 y+x=4(x+y)+3$

$\Rightarrow 10 y+x=4 x+4 y+3$

$\Rightarrow 4 x+4 y-10 y-x=-3$

$\Rightarrow 3 x-6 y=-3$

$\Rightarrow 3(x-2 y)=-3$

$\Rightarrow x-2 y=-\frac{3}{3}$

$\Rightarrow x-2 y=-1$

After interchanging the digits, the number becomes $10 x+y$.

If 18 is added to the number, the digits are reversed. Thus, we have

$(10 y+x)+18=10 x+y$

$\Rightarrow 10 x+y-10 y-x=18$

$\Rightarrow 9 x-9 y=18$

$\Rightarrow 9(x-y)=18$

$\Rightarrow x-y=\frac{18}{9}$

$\Rightarrow x-y=2$

So, we have the systems of equations

$x-2 y=-1$

$x-y=2$

Here x and y are unknowns. We have to solve the above systems of equations for x and y.

Subtracting the first equation from the second, we have

$(x-y)-(x-2 y)=2-(-1)$

$\Rightarrow x-y-x+2 y=3$

$\Rightarrow y=3$

Substituting the value of in the first equation, we have

$x-2 \times 3=-1$

$\Rightarrow x-6=-1$

$\Rightarrow x=-1+6$

$\Rightarrow x=5$

Hence, the number is $10 \times 3+5=35$.