Question

the digits of two numbers differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What is the original number ​

Answer 1

Answer:

Let the number be xy

Y= x-3

Number formed:

10(x)+ x-3 [x will be multiplied by 10 because of its place]

10x+(x-3)

Reversed number= yx

= 10(x-3)+x [x-3 will be multiplied by 10 because of its place]

= 10x-30+x

Sum of the numbers= 143

10x+(x-3)+10x-30+x= 143

11x+11x-3-30= 143

22x-33= 143

22x= 143+33

22x= 176

X= 176/22

X= 8

NUMBERS FORMED:

xy= 58

Yx= 85

VERIFY:

The sum of both the numbers should be equal to 143. Using this statement, we can verify our answer.

Xy+yx= 143

58+85= 143

143= 143

LHS= RHS

Hence, verified

Hope it helps

Please mark my answer as BRAINLIEST

Answer 2

Answer:

$\huge\mathcal{Question:}$

The digits of two numbers differ by 3. If the digits are interchanged and the resulting number is added to the original number, we get 143. What is the original number .

$\huge\mathcal{Answer:}$

Let the two digits be x and y (where x is at tens place and y is at once place)

so the number we get is 10x+y

x - y = 3 .....(given in the Question) -----eq1

Now interchanging the digits , we get

The new number is 10y+x because now x is at once place and y is at the tens place.

now,

old number + new number = 143 (given). -----eq2

now putting the values,

10x+y+10y+x = 143

$\implies$11x + 11 y = 143

$\implies$11(x+y) = 143

$\implies$x+y = $\sf\frac{143}{11}$

$\implies$x+y = 13

now from eq1 ,we come to know that y = 3+x

so putting y = 3x , we get

x+(x+3) = 13

$\implies$2x+3 = 13

$\implies$2x = 10

$\implies$x = 5

so now from y = x+3 , we get the value of y to be 8.

So the original number is 10×5 + 8 =58

$\boxed{Answer=58}$