Question

the sum of digits of twoo digit number is 7 if the digits are reserved the new number decreased by 2 equals twice the original number find the number​

Answer 1

Solution :-

Let :-

Digit in ten's place = x

Digit in one's place = y

Two digit number = 10x + y

According to the first condition :-

$\longrightarrow \sf x+y = 7 \\\\\longrightarrow x = 7-y \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(1)$

According to the second condition :-

$\longrightarrow \sf 2(10x+y) = 10y+x-2 \\\\\longrightarrow 20x+2y = 10y+x -2 \\\\\longrightarrow 20x-x+2y-10y = -2 \\\\\longrightarrow 19x-8y = -2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ...............(2)$

Substitute the value of x in eq(2) :-

$\longrightarrow \sf 19(7-y)-8y = -2 \\\\\longrightarrow 133-19y-8y = -2 \\\\\longrightarrow -27y = -135 \\\\\longrightarrow \bf y = 5$

Substitute y = 5 in eq(1) :-

$\longrightarrow \sf x = 7-5 \\\\\longrightarrow \bf x=2$

Two digit number = 25

Answer 2

Step-by-step explanation:

Solution :-

Let the digits be x and y

so,the original number be 10x + y

(since x in tenth place and y in unit place)

Also,

Let Digit in ten's place = x

Let Digit in one's place = y

Two digit number = 10x + y

The first condition gives:

=> x+y = 7

=> x = 7-y ...............(1)

The second condition gives:

=>2(10x+y) = 10y+x-2

=> 20x+2y = 10y+x -2

=> 20x-x+2y-10y = -2

=> 19x-8y = -2 ...............(2)

Thus,we have two equations

Substitute the value of x in eq(2) :-

19(7-y)-8y = -2

133-19y-8y = -2

-27y = -135

y = 5

Substitute y = 5 in eq(1) :-

x = 7-5

x=2

Therefore,

original number is 25