Verified answer

[tex]\huge{\fcolorbox{maroon}{s} {\large{\fcolorbox{white}{mistyrose}{{\fcolorbox{gold} {White}{Answer}}}}}}[/tex]

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Let the digit at tens place be x and at ones place be y.

Two digit number = [tex] 10x + y [/tex]

Reversed order of number = [tex] 10y + x [/tex]

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Given that,

▪Sum of the digits of the number = 9

[tex] So, \: x + y = 9\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ......(1) \\ [/tex]

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▪Nine times the number is twice the reversed order of number.

[tex] So, \: 9(10x+y) = 2(10y+x) [/tex]

[tex] \hookrightarrow 90x + 9y = 20y × 2x [/tex]

[tex] \hookrightarrow 90x - 2x = 20y - 9y [/tex]

[tex] \hookrightarrow 88x = 11y [/tex]

[tex] \hookrightarrow 88x - 11y = 0 [/tex]

[tex] \hookrightarrow 11(8x - y) =0 [/tex]

[tex] \hookrightarrow (8x - y) = \frac{0}{11} \\ [/tex]

[tex] \hookrightarrow (8x - y) = 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ......(2) [/tex]

▪Hence our equations are :--

[tex] x + y = 9\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ......(1) \\ [/tex]

[tex] (8x - y) = 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ......(2) \\ [/tex]

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▪Using elimination method with equations by adding them ;

[tex] x \: \: \cancel{ + y} = 9 \\ \underline{ 8x \: \cancel{- y} = 0} \\ \underline{ \: 9x \: \: \: = \: \: \: 9} \\ [/tex]

[tex] \dashrightarrow x = \frac{9}{9} \\ [/tex]

[tex] \dashrightarrow x = 1 [/tex]

▪Substituting the value of x in eq. 1 ;

[tex] \hookrightarrow 1 + y = 9 [/tex]

[tex] \hookrightarrow y = 9-1 [/tex]

[tex] \hookrightarrow y = 8 [/tex]

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☆ As x = 1 and y = 8,

▪So, the required number is [tex] 10x+y [/tex]

[tex] \hookrightarrow 10(1) + 8 [/tex]

[tex] \hookrightarrow 10 + 8 [/tex]

[tex] \hookrightarrow \boxed{\purple{\bm{18}}} [/tex]

☆ Therefore, the required number is 18.

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