Given : [tex]\bf{Equation}\begin{cases}\sf{2x – y = 5\:\qquad \longrightarrow \bf{Eq.1}}\\\\\sf{3x + y = 10\:\qquad \longrightarrow \bf{Eq.2}}\end{cases}\\\\[/tex] Need To Find : The Value of x and y . ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ [tex]\qquad \qquad \bigstar \sf{2x – y = 5\:\qquad \longrightarrow \bf{Eq.1}}\\[/tex] [tex]\qquad \qquad \bigstar \sf{3x + y = 10\:\qquad \longrightarrow \bf{Eq.1}}\\[/tex] ⠀⠀⠀⠀⠀⠀[tex]\underline {\sf{\star\:Now \: By \: Using\:\bf{Elimination \:method}\: with \: both \: Equation \::}}\\[/tex] [tex]\boxed{\begin {array}{ccc}\\ \dag\quad \large\underline{\bf Elimination\:Method\::}\\ \\ \sf{ 2x\:\: } \sf{\cancel{-y}\:\:} \sf{=5}\\ \\\sf{ 3x\:\: } \sf{\cancel{+y}\:\:} \sf{=10}\\\\ \underline {\qquad \quad}\\\\\sf{5x\:\:}\sf{\:\:\:\:}\sf{=5}\end{array}}[/tex] [tex]:\implies \sf{ 5x = 15 }\\\\:\implies \sf{ x = \cancel {\dfrac{15}{3}}}\\\\\underline {\boxed{\pink{ \mathrm { x = 3\: }}}}\:\bf{\bigstar}\\[/tex] ⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Substituting \: \bf{x = 3} \: in \: Eq.1 \::}}\\[/tex] [tex]\qquad \qquad \bigstar \sf{2x – y = 5\:\qquad \longrightarrow \bf{Eq.1}}\\[/tex] [tex] :\implies \sf{ 2 \times 3 – y = 5 }\\\\:\implies \sf{ 6 – y = 5 }\\\\:\implies \sf{ 2 \times 3 – y = 5 – 6 }\\\\:\implies \sf{\cancel {-} y = \cancel{-}1 }\\\\\underline {\boxed{\pink{ \mathrm { y = 1\: }}}}\:\bf{\bigstar}\\[/tex] Therefore, ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm { The\:Value \:of\:x \:is\:\bf{3\: }}}}\\[/tex] And , ⠀⠀⠀⠀⠀[tex]\underline{ \mathrm { The\:Value \:of\:y \:is\:\bf{1\: }}}\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
Given : [tex]\bf{Equation}\begin{cases}\sf{2x – y = 5\:\qquad \longrightarrow \bf{Eq.1}}\\\\\sf{3x + y = 10\:\qquad \longrightarrow \bf{Eq.2}}\end{cases}\\\\[/tex]
Need To Find : The Value of x and y .
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[tex]\qquad \qquad \bigstar \sf{2x – y = 5\:\qquad \longrightarrow \bf{Eq.1}}\\[/tex]
[tex]\qquad \qquad \bigstar \sf{3x + y = 10\:\qquad \longrightarrow \bf{Eq.1}}\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\sf{\star\:Now \: By \: Using\:\bf{Elimination \:method}\: with \: both \: Equation \::}}\\[/tex]
[tex]\boxed{\begin {array}{ccc}\\ \dag\quad \large\underline{\bf Elimination\:Method\::}\\ \\ \sf{ 2x\:\: } \sf{\cancel{-y}\:\:} \sf{=5}\\ \\\sf{ 3x\:\: } \sf{\cancel{+y}\:\:} \sf{=10}\\\\ \underline {\qquad \quad}\\\\\sf{5x\:\:}\sf{\:\:\:\:}\sf{=5}\end{array}}[/tex]
[tex]:\implies \sf{ 5x = 15 }\\\\:\implies \sf{ x = \cancel {\dfrac{15}{3}}}\\\\\underline {\boxed{\pink{ \mathrm { x = 3\: }}}}\:\bf{\bigstar}\\[/tex]
⠀⠀⠀⠀⠀⠀[tex]\underline {\frak{\star\:Now \: By \: Substituting \: \bf{x = 3} \: in \: Eq.1 \::}}\\[/tex]
[tex]\qquad \qquad \bigstar \sf{2x – y = 5\:\qquad \longrightarrow \bf{Eq.1}}\\[/tex]
[tex] :\implies \sf{ 2 \times 3 – y = 5 }\\\\:\implies \sf{ 6 – y = 5 }\\\\:\implies \sf{ 2 \times 3 – y = 5 – 6 }\\\\:\implies \sf{\cancel {-} y = \cancel{-}1 }\\\\\underline {\boxed{\pink{ \mathrm { y = 1\: }}}}\:\bf{\bigstar}\\[/tex]
Therefore,
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm { The\:Value \:of\:x \:is\:\bf{3\: }}}}\\[/tex]
And ,
⠀⠀⠀⠀⠀[tex]\underline{ \mathrm { The\:Value \:of\:y \:is\:\bf{1\: }}}\\[/tex]
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