Question

the sum of the digit of a two digit number is 3 .if 9 is added to the number the digits are reversed .find the number .

Answer 1

HELLO!

LeT the two digits be x and y.

x+y = 3 …(1)

10x+y+9 = 10y+x, or

9x=9y=9, or

x-y=1 …(2)

Add (1) and (2)

2x=4 or x = 2,ad so y = 1

So the number is 21.

keep smiling

:)

Answer 2

Answer:

$\large\bf\underline\red{Question➡} \\ \\\bf \:the \: sum \: of \: a \: two \: digit \: \\\bf number \: is \: 3. \: if \: 9 \: is \: added \: to \: \\\bf the \: number \: its \: digit \: are \: \\ \bf \: reversed \: . \: then \: find \: the \: \\\bf number \: \\ \\ \large\bf\underline\red{answer➡} \: \\ \\ \large{\boxed{\mathfrak\red{\fcolorbox{magenta}{aqua}{21}}}} \\ \\ \large\bf\underline\red{solution➡} \\ \\ \sf \: let \: number \: be \: \bf \: x \:\sf and \: \bf \: y \\ \\ \bf\underline{according \: to \: the \: question} \\ \\ \sf \: x + y \: = 3 \: ......\bf \: eq.(1) \\ \\ \sf 10x + y + 9 = 10y + x \\ \\ \sf 9x - 9y = 9 \\ \\ \sf x - y = 1 \: ......\bf \: eq.(2) \\ \\ \\ \bf\underline{add \: eq.(1) \: and \: (2)} \\ \\ \sf \: x + y \: = 3 \\ \sf x - y = 1 \\ \\ \sf 2x = 4 \\ \\ \sf x = \frac{4}{2} \\ \\ \bf\blue{ x = 2 }\\ \\ \bf\underline{put \: the \: value \: of \: x \: in \: eq.(1)} \\ \bf\underline{we \: get \:➡ } \\ \\ \bf\blue{y = 1} \\ \\ \bf\underline{our \: required \: answer \: is \:➡ } \\ \\ \large{\boxed{\mathfrak\red{\fcolorbox{magenta}{aqua}{21}}}}$