Question:-
The sum of the squares of the digits of a two-digit number is 13.If we subtract 9 from that number , we get a num

2 thoughts on “Question:-<br /> The sum of the squares of the digits of a two-digit number is 13.If we subtract 9 from that number , we get a num”

1. Acçórding to the quéstion:-

   ⚘Súbtract 9 fróm the oríginál númber , we get a numbér consísting of the sáme digits writtén in the revérse órder.

   [tex] \sf{ \to \color{blue}óriginal\:number-9 = réverse\: fórm\:óf \: óriginàl\:númbér}[/tex]

   [tex] \sf{ \implies(10y+ a) – 9 = (10a +y)}[/tex]

   Tránspósíng the líke térms

   [tex] \sf{ \implies10y -y -9 = 10a-a} \\ \\ \sf{ \implies9y -9 = 9a} \\ \\ \sf{ \implies9×(y-1 )= 9a} \\ \\ \bold{ \implies \red{y -1 = x}\: \: \: \: ….(2.)}[/tex]

   Nów súbsitúte the válúe of x = ( y-1) in a ²+y² = 13

   [tex] \sf{ \implies a²+y² = 13} \\ \\

   \sf{ \implies(y-1)² +y² =13 }\\ \\

   \sf{ \implies y²+1 -2y +y² =13 }\\ \\

   \sf{ \implies 2y² -2y +1 =13 }\\ \\

   \sf{ \implies 2y² -2y = 12 }\\ \\

   y²- y = 6[/tex]

   míddlé splít the quàdratiç éqúatión

   [tex] \sf{ \implies y²+2y -3y -6 = 0 }\\ \\ \sf{ \implies y (y+2)-3(y+2) = 0} \\ \\ \sf{\implies(y+2)(y-3) = 0 }\\ \\ \sf{ \implies y = – 2 \: ór \: 3}[/tex]

   ⚘So the válue of y is 3

   pút the valúe of y in eqúatíon (2)

   [tex] \sf{ \implies x = y-1 }\\ \\ \sf{ \implies a= 3-1 }\\ \\ \sf{ \implies \red{ a= 2}}[/tex]

   ⚘Theréfóre the númbér fórméd is 32

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2. Question:-

   • The sum of the squares of the digits of a two-digit number is 13.If we subtract 9 from that number , we get a number consisting of the same digits written in the reverse order. Find the number.

   To Find:–

   • Fine the number.

   Solution:–

   Here ,

   Let the numbers be x and y

   Given ,

   Sun of the squares of number is 13:-

   • [tex]\tt \: { x }^{ 2 } + { y }^{ 2 } = 13 [/tex]

   According to the Question:-

   [tex]\tt\implies \: 10x + y – 9 = 10y + x [/tex]

   [tex]\tt\implies \: 10x – x – 10y + y = 9 [/tex]

   [tex]\tt\implies \: 9x – 9y = 9 [/tex]

   [tex]\tt\implies \: 9( x – y ) = 9 [/tex]

   [tex]\tt\implies \: x – y = \cancel\dfrac { 9 } { 9 } [/tex]

   [tex]\tt\implies \: x – y = 1 . . . . ( i ) [/tex]

   Now ,

   [tex]\tt\implies \: { ( x – y ) }^{ 2 } = { x }^{ 2 } + { y }^{ 2 } – 2xy [/tex]

   [tex]\tt\implies \: 1 = 13 – 2xy [/tex]

   [tex]\tt\implies \: 2xy = 12 [/tex]

   [tex]\tt\implies \: xy = \cancel\dfrac { 12 } { 2 } [/tex]

   [tex]\tt\implies \: y = \dfrac { 6 } { x } [/tex]

   Now ,

   Substitute y in equation ( i ):-

   [tex]\tt\implies \: x – \dfrac { 6 } { x } = 1 [/tex]

   [tex]\tt\implies \: { x }^{ 2 } – x – 6 = 0 [/tex]

   [tex]\tt\implies \: { x }^{ 2 } + 2x – 3x – 6 = 0 [/tex]

   [tex]\tt\implies \: x( x + 2 ) – 3( x + 2 ) = 0 [/tex]

   [tex]\tt\implies \: ( x + 2 ) ( x – 3 ) = 0 [/tex]

   [tex]\tt\implies \: x = – 2 , 3 [/tex]

   Hence ,

   • The number is 32

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