[tex] {( \sqrt{3} )}^{x + y} = 9 \\ {( \sqrt{2} )}^{x – y} = 32[/tex] find the value of 2x + y.​

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[tex] {( \sqrt{3} )}^{x + y}  = 9 \\ {( \sqrt{2} )}^{x – y}  = 32[/tex]

find the value of 2x + y.​

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Doris 1 year 2021-09-03T15:08:25+00:00 1 Answers 5 views 0

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    2021-09-03T15:10:08+00:00

    [tex] \bf \purple{ \underline{Given :-}}[/tex]

    [tex] • \:  {( \sqrt{3} )}^{x + y}  = 9\:  \:  \: \: (i)[/tex]

    [tex]• \:  {( \sqrt{2} )}^{x – y}  = 32 \:  \:  \: \: (ii)[/tex]

    [tex] \\ [/tex]

    [tex]  \bf \purple{ \underline{To  \: Find :- }}[/tex]

    [tex]  •   {\sf{The  \: value  \: of}} \:   \: 2x+ y.[/tex]

    [tex] \\ [/tex]

    [tex] \huge\bf \purple{ \underline{Solution :- }}[/tex]

    [tex]  \sf{From \:  equation \:  (i), }[/tex]

    [tex] {( \sqrt{3} )}^{x + y}  = 9 [/tex]

    [tex]⇒ {( \sqrt{3} )}^{x + y}  =  ({ \sqrt{3} })^{4} [/tex]

    [tex] ⇒ x + y = 4 \:  \:  \: \: (iii)[/tex]

    [tex] \\ [/tex]

    [tex] \sf{From \:  the  \: equation \:  (ii)}[/tex]

    [tex]{( \sqrt{2} )}^{x – y}  = 32[/tex]

    [tex] ⇒( { \sqrt{2} })^{x – y}  =(  { \sqrt{2} })^{10} [/tex]

    [tex]⇒x – y = 10 \:  \:  \: \: (iv)[/tex]

    [tex] \\ [/tex]

    [tex]\sf{We \:  have \:  to  \: add \:  equation \:  (iii)  \: and \:  equation  \: (iv)}[/tex]

    [tex] x + y + x – y = 4 + 10[/tex]

    [tex] ⇒2x = 14[/tex]

    [tex] ⇒x = 7[/tex]

    [tex] \\ [/tex]

    [tex] \sf{Subtracting \:  equation \:  (iii)  \: from  \: equation  \: (ii),  }[/tex]

    [tex]x  – y – x – y = 10 – 4[/tex]

    [tex]⇒ – 2y = 6[/tex]

    [tex]⇒y =  – 3[/tex]

    [tex] \\ [/tex]

    [tex] \bf\therefore x = 7 \: \:  and  \: \: y =  – 3[/tex]

    [tex] \\ [/tex]

    [tex] { \sf{Th e \:  value \:  of }}= 2x +y [/tex]

    [tex]\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: \: \: \: \: =2 \times 7  + ( – 3)[/tex]

    [tex]\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: \: \: \: \: =14 – 3[/tex]

    [tex] \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \: \: \: \: =11[/tex]

    [tex] \bf \red{Hence,  \: the  \: value \:  of  \: 2x + y  \: is  \: 11. }[/tex]

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