Seven times a two-digit number is equal to four times the number obtained by reversing the digits.

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 difference between the two digits of the number is 3 . Thus, we have $x-y=\pm 3$

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

Seven times the number is equal to four times the number obtained by reversing the order of the digits. Thus, we have

$7(10 y+x)=4(10 x+y)$

$\Rightarrow 70 y+7 x=40 x+4 y$

$\Rightarrow 40 x+4 y-70 y-7 x=0$

$\Rightarrow 33 x-66 y=0$

$\Rightarrow 33(x-2 y)=0$

$\Rightarrow x-2 y=0$

So, we have two systems of simultaneous equations

$x-y=3$,

$x-2 y=0$

$x-y=-3$,

 

$x-2 y=0$

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

(i) First, we solve the system

$x-y=3$

$x-2 y=0$

Multiplying the first equation by 2 and then subtracting from the second equation, we have

$(x-2 y)-2(x-y)=0-2 \times 3$

$\Rightarrow x-2 y-2 x+2 y=-6$

$\Rightarrow-x=-6$

 

$\Rightarrow x=6$

Substituting the value of $x$ in the first equation, we have

$6-y=3$

$\Rightarrow y=6-3$

 

$\Rightarrow y=3$

Hence, the number is $10 \times 3+6=36$.

(ii) Now, we solve the system

$x-y=-3$,

 

$x-2 y=0$

Multiplying the first equation by 2 and then subtracting from the second equation, we have

$(x-2 y)-2(x-y)=0-(-3 \times 2)$

$\Rightarrow x-2 y-2 x+2 y=6$

$\Rightarrow-x=6$

 

$\Rightarrow x=-6$

Substituting the value of x in the first equation, we have

$-6-y=-3$

$\Rightarrow y=-6+3$

$\Rightarrow y=-3$

But, the digits of the number can’t be negative. Hence, the second case must be removed.