Question

sum of 2 digit is 9. If digits are reserved , the no. is 63 more than the original no. find the no.

Answer 1

Let x be the ones digit
⇒ The tens digit is 9−x

10(9−x)+x+63=10x+(9−x)

⇒90−10x+x+63=10x+9−x

⇒153−9x=9x+9

⇒153−9=18x

⇒144=18x
⇒x=8

⇒9−x=9−8=1

Therefore, the number we are looking for is

10(1)+8=18

Answer 2

Hey mate!

Here's your answer!!



Let x be the ones digit.

The ten's digit is 9 - x.

➡ 10 (9 - x) + x + 63 = 10x + (9 - x)

➡ 90 - 10x + x + 63 = 10x + 9 - x

➡ 153 - 9x = 9x + 9

➡ 153 - 9 = 18x

➡ 144 = 18x

➡ x = 8

➡ 9 - x = 9 - 8 = 1

Therefore, the number we are looking for is
10 (1) + 8 = 18.

$hope \: it \: helps \: you....$
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