Question

plz give the correct anawer​

Answer 1

Answer:

maithili

1s upon multiple 1st equation by 5 and 2nd equation by 2 we get

Step-by-step explanation:

multiple 1st equation by 5 and 2nd equation by 2 we get

and after solving X is eliminate

Answer 2

$\bf{Given \::}\begin {cases} &\sf{2x - 3y = 14 \qquad \longrightarrow\:\:Eq.1}\\\\&\sf{ 5x + 2y = 16 \qquad \longrightarrow \:\:Eq.2}\end{cases}$

Exigency To Solve : The Given Simultaneous Equation [ Eq.1 & Eq.1 ] by Elimination method .

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$\qquad \star\:\:\bf{ \bigg( 2x - 3y = 14\bigg)\qquad \longrightarrow \:\:Eq.1 }$

$\qquad \star\:\:\bf{ \bigg( 5x + 2 y = 16\bigg)\qquad \longrightarrow \:\:Eq.2}$

⠀⠀⠀⠀⠀⠀⠀⠀We have to solve given Simultaneous Equation using Elimination method.

⠀⠀⠀⠀ In Elimination method we have to eliminate one variable for eliminating one variable we have to have one variable equal in both Equation . So , for this ,

• Multiply Eq.1 by 2 and Eq.2 by 3 .

⠀⠀⠀⠀ Multiplying Eq.1 by 2 .

$\qquad :\implies \:\:\sf{ \bigg( 2x - 3y = 14\bigg)\qquad \longrightarrow \:\:Eq.1 }$

$\qquad :\implies \:\:\sf{ \bigg( 2x - 3y = 14\bigg)\times 2 }$

$\qquad :\implies \:\:\bf{ \bigg( 4x - 6y = 28\bigg)\qquad \longrightarrow Eq.1 }$

⠀⠀⠀⠀Multiplying Eq.2 by 3 .

$\qquad :\implies \:\:\sf{ \bigg( 5x + 2 y = 16\bigg)\qquad \longrightarrow \:\:Eq.2}$

$\qquad :\implies \:\:\sf{ \bigg( 5x + 2 y = 16\bigg)\times 3}$

$\qquad :\implies \:\:\sf{ \bigg( 15x + 6 y = 48\bigg)\qquad\longrightarrow\:\:Eq.2}$

Therefore,

• $\qquad :\implies \:\:\bf{ \bigg( 4x - 6y = 28\bigg)\qquad \longrightarrow Eq.1 }$
• $\qquad :\implies \:\:\sf{ \bigg( 15x + 6 y = 48\bigg)\qquad\longrightarrow\:\:Eq.2}$

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Eliminating \: one \: Variable \: from\:Eq.1 \:\&\:Eq.2 \::}}$

$\boxed{\begin {array}{ccc}\\ \dag\quad \underline{\bf Elimination \:Method:-}\\ \\ \sf{4x - 6y = 28 }\\ \\ \sf{15 x + 6y = 48 }\\ \\ \underline {\qquad \quad}\\ \\ \sf{19 x \:\:= 76}\end{array}}$

$\qquad :\implies \:\:\sf{ 19x = 76}$

$\qquad :\implies \:\:\sf{ x =\cancel {\dfrac{76}{19}}}$

$\qquad :\implies \:\:\bf{ x = 4}$

⠀⠀⠀⠀⠀⠀$\underline {\sf{\star\:Now \: By \: Substituting \: x =4 \: in \: Eq.1 \::}}$

$\qquad :\implies \:\:\sf{ \bigg( 2x - 3y = 14\bigg)\qquad \longrightarrow \:\:Eq.1 }$

$\qquad :\implies \:\:\sf{ 2\times 4 - 3y = 14 }$

$\qquad :\implies \:\:\sf{ 8 - 3y = 14 }$

$\qquad :\implies \:\:\sf{ - 3y = 14 - 8 }$

$\qquad :\implies \:\:\sf{ - 3y = 6 }$

$\qquad :\implies \:\:\sf{ y = \cancel {\dfrac{6}{-3}}}$

$\qquad :\implies \:\:\bf{ y = -2 }$

Therefore,

⠀⠀⠀⠀⠀$\therefore \underline{ \mathrm { Hence \:The\:Value \:of\:x \:is :\bf{4\: }\:\sf{\&\:Value\:of\:y \:is }\:\bf{-2}\:\:}}$

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