ANSWER:

The price of a smartphone is 5000 more than the cost of two feature phones &

the cost of 4 feature phones and two smartphone is 88,000 .

Exigency To Find : The cost of a smartphone and a feature phone.

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❍ Let's Consider the cost price of smart phone and feature phone be Rs. x & Rs. y , respectively .

⠀⠀⠀⠀⠀⠀CASE I : The price of a smartphone is Rs. 5000 more than the cost of two feature phones .

\begin{gathered}\qquad :\implies \sf x = 2y + 5000 \:\:\\\end{gathered}

:⟹x=2y+5000

\begin{gathered}\qquad :\implies \bf x = 2y + 5000 \:\:\:\qquad\qquad \bigg\lgroup \sf{ Eq^n \: 1 }\bigg\rgroup\\\end{gathered}

:⟹x=2y+5000

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Eq

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⠀⠀⠀⠀⠀⠀CASE II : The cost of 4 feature phones and two smartphones is Rs. 88,000 .

\begin{gathered}\qquad :\implies \sf 4y + 2x = 88000 \:\:\\\end{gathered}

:⟹4y+2x=88000

\begin{gathered}\qquad :\implies \bf 4y + 2x = 88000 \:\:\:\qquad\qquad \bigg\lgroup \sf{ Eq^n \: 2}\bigg\rgroup\\\end{gathered}

:⟹4y+2x=88000

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Eq

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⠀⠀⠀⠀⠀⠀Now , Finding the cost of Smart phone & Feature phone :

\begin{gathered}\qquad \:\maltese\:\bf{ From \:EQUATION \:2 \:\::}\\\end{gathered}

✠FromEquation2:

\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:2\:: 4y + 2x = 88000 }\bigg\rgroup \\\\\end{gathered}

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Equation2:4y+2x=88000

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\begin{gathered}\qquad :\implies \sf 4y + 2x = 88000 \:\:\\\end{gathered}

:⟹4y+2x=88000

⠀⠀⠀⠀⠀⠀\begin{gathered}\UNDERLINE {\boldsymbol{\star\:Now \: By \: Substituting \: the \: Eq^n\:1 \: : \::}}\\\end{gathered}

⋆NowBySubstitutingtheEq

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\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup \\\\\end{gathered}

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Equation1:x=2y+5000

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\begin{gathered}\qquad :\implies \sf 4y + 2x = 88000 \:\:\\\end{gathered}

:⟹4y+2x=88000

\begin{gathered}\qquad :\implies \sf 4y +2 (2y + 5000) = 88000 \:\:\\\end{gathered}

:⟹4y+2(2y+5000)=88000

\begin{gathered}\qquad :\implies \sf 4y +4y + 10000 = 88000 \:\:\\\end{gathered}

:⟹4y+4y+10000=88000

\begin{gathered}\qquad :\implies \sf 4y +4y = 88000 - 10000 \:\:\\\end{gathered}

:⟹4y+4y=88000−10000

\begin{gathered}\qquad :\implies \sf 8y = 88000 - 10000 \:\:\\\end{gathered}

:⟹8y=88000−10000

\begin{gathered}\qquad :\implies \sf 8y = 78000 \:\:\\\end{gathered}

:⟹8y=78000

\begin{gathered}\qquad :\implies \sf y = \dfrac{ 78000}{8} \:\:\\\end{gathered}

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78000

\begin{gathered}\qquad :\implies \sf y = \cancel {\dfrac{ 78000}{8}} \:\:\\\end{gathered}

:⟹y=

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78000

\begin{gathered}\qquad :\implies \bf y = 9750 \:\:\\\end{gathered}

:⟹y=9750

\begin{gathered}\qquad :\implies \frak{\underline{\PURPLE{\:y = Rs.\: 9750 }} }\:\:\bigstar \\\end{gathered}

:⟹

y=Rs.9750

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⠀⠀⠀⠀⠀⠀\begin{gathered}\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: \: Value\:of\: y \ [ \ 9750 \ ] \:in \:Eq^n \:1 \::}}\\\end{gathered}

⋆NowBySubstitutingtheValueofy [ 9750 ]inEq

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\begin{gathered}\qquad \dag\:\:\bigg\lgroup \sf{Equation \:1\:: x = 2y + 5000 }\bigg\rgroup \\\\\end{gathered}

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Equation1:x=2y+5000

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\begin{gathered}\qquad :\implies \sf x = 2y + 5000 \:\:\\\end{gathered}

:⟹x=2y+5000

\begin{gathered}\qquad :\implies \sf x = 2(9750) + 5000 \:\:\\\end{gathered}

:⟹x=2(9750)+5000

\begin{gathered}\qquad :\implies \sf x = 19500 + 5000 \:\:\\\end{gathered}

:⟹x=19500+5000

\begin{gathered}\qquad :\implies \bf x = 24500 \:\:\\\end{gathered}

:⟹x=24500

\begin{gathered}\qquad :\implies \frak{\underline{\purple{\:x \:= Rs.\: 24500 }} }\:\:\bigstar \\\end{gathered}

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x=Rs.24500

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Therefore,

The cost of smartphone is : x = Rs. 24, 500

The cost of feature phone: y = Rs. 9,750

Therefore,

⠀⠀⠀⠀⠀\begin{gathered}\qquad \therefore {\underline{ \sf \:Cost \:of\:Smartphone \:and \:feature \:phone \:are\:\bf Rs. \ 24,500 \:\& \: Rs.\: 9,750\:\:\sf , \ respectively . }}\\\end{gathered}

∴

CostofSmartphoneandfeaturephoneareRs. 24,500&Rs.9,750, respectively.

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