Question

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Answer 1

Answer:

$\begin{gathered}{\underline{\underbrace{\Huge{\bigstar{\textsf{\textbf{\blue{Solution :}}}}}}}}\end{gathered}$

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${\large \qquad\boxed{{\begin{array}{cc} \tt Let \: Length \: of \: Rectangle \: be \: x \: Units \\ \tt \& \: Breadth \: of \: rectangle \: be \: y \:units\end{array}}}}$

$\tt Hence,$

$\tt Area = Length × Breadth$

$\tt Area = xy$

$\tt \underline{Given,}$

${\large \qquad\boxed{{\begin{array}{cc} \tt Area \: Gets \: Redused \: by \: 9 \: square \: units \\ \\ \tt Length \: is \: Reduced \: by \: 5 \: units \\ \\ \tt Breadth \: increased \: by \: 3 \: units \end{array}}}}$

$\tt New \: area = New \: Length × New \: Breadth$

$\tt Old \: Area -9 =(Length -5)×(Breadth+3)$

$\orange{\large \boxed{\boxed{\begin{array}{cc}\footnotesize \tt{xy - 9 = (x - 5)(y + 3)} \\ \\\footnotesize \tt{ xy - 9 = x(y + 3) - 5(y + 3)} \\ \\ \footnotesize \tt{xy - 9 = xy + 3x - 5y - 15} \\ \\ \footnotesize \tt{0 = xy + 3x - 5y - 15 - xy + 9} \\ \\ \footnotesize \tt{3x - 5y = 6}\xrightarrow{\text{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}(1)\end{array}}}}$

$\tt \underline{So,}$

$\tt New \: area = New \: Length × New \: Breadth$

$\tt Old \: Area + 67 =(Length + 3)×(Breadth+2)$

$\pink{\large \boxed{\boxed{\begin{array}{cc} \footnotesize \tt{xy + 6y = (x + 3)(y + 2)} \\ \\ \footnotesize \tt{xy + 67 = x(y + 2) + 3(y + 2)} \\ \\ \footnotesize \tt{xy + 67 = xy + 2x+ 3y + 6} \\ \\ \footnotesize \tt{0 = xy + 2x + 3y + 6 - xy - 67} \\ \\ \footnotesize \tt{2x + 3y - 61 = 0} \\ \\ \footnotesize \tt{2x + 3y = 61\xrightarrow{\text{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}(2)} \end{array}}}}$

${\large\boxed{{\begin{array}{cc} \tt Here \: Our \: Equations \: Are \\ \\ \footnotesize \tt{3x - 5y = 6}\xrightarrow{\text{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}(1) \\ \footnotesize \tt{2x + 3y = 61\xrightarrow{\text{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}(2)} \end{array}}}}$

$\tt{From(1)}$

${\large \boxed{\boxed{\begin{array}{cc} \footnotesize \tt{3x - 5y - 6 = 0} \\ \\ \footnotesize \tt{3x = 6x + 5y} \\ \\ \footnotesize \tt{x = \dfrac{6 + 5y}{3}} \end{array}}}}$

$\tt{Putting \: Value \: of \: x \: in (2)}$

${\large \boxed{\boxed{\begin{array}{cc} \footnotesize \tt{2x + 3y = 61} \\ \\ \footnotesize \tt{2 \bigg( \dfrac{6 + 5y}{3} \bigg) + 3y = 61} \\ \\ \end{array}}}}$

$\tt Multiplying \: both \: sides \: by \: 3$

$\purple{\large \qquad \boxed{\boxed{\begin{array}{cc} \qquad \footnotesize \tt{3 \times 2 \bigg( \dfrac{(6 + 5y)}{3} \bigg) + 3 \times 3y = 3 \times 61} \\ \\\qquad \footnotesize \tt{2(6 + 5y) + 9y = 183} \\ \\\qquad \footnotesize\tt{12 + 10y + 9y = 183} \\ \\ \qquad\footnotesize \tt{19y = 183 - 12} \\ \\ \qquad \footnotesize \tt{19y = 171} \\ \\ \qquad \footnotesize \tt{y = \cancel \dfrac{171}{9} } \\ \\ \qquad \footnotesize \tt{y = 9} \\ \\ \footnotesize \tt{Putting \: y=9 \: in \: Equation (1)} \\ \\ \qquad \footnotesize \tt{3x - 5y = 6} \\ \\ \qquad \footnotesize \tt{3x - 5(9) = 6} \\ \\ \qquad \footnotesize \tt{3x - 45= 6} \\ \\ \qquad \footnotesize \tt{3x = 6 + 45} \\ \\ \qquad \footnotesize \tt{3x = 51} \\ \\ \qquad \footnotesize \tt{x = \cancel\dfrac{51}{3} } \\ \\ \qquad \footnotesize \tt{x = 17} \end{array}}}}$

$\tt{Therefore \: x=17,y=9}$

$\orange{\large\boxed{{\begin{array}{cc} \red{\begin{cases} \tt Length \: of \: Rectangle = x =17 \: Units \\  \\  \tt Breadth \: of \: Rectangle = y= 9 \: units\end{cases}} \end{array}}}}$

$\purple{\rule{45pt}{7pt}}\red{\rule{45pt}{7pt}}\pink{\rule{45pt}{7pt}}\blue{\rule{45pt}{7pt}}$

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Answer 2

$\begin{gathered}\purple{\large \qquad \boxed{\boxed{\begin{array}{cc} \qquad \footnotesize \tt{3 \times 2 \bigg( \dfrac{(6 + 5y)}{3} \bigg) + 3 \times 3y = 3 \times 61} \\ \\\qquad \footnotesize \tt{2(6 + 5y) + 9y = 183} \\ \\\qquad \footnotesize\tt{12 + 10y + 9y = 183} \\ \\ \qquad\footnotesize \tt{19y = 183 - 12} \\ \\ \qquad \footnotesize \tt{19y = 171} \\ \\ \qquad \footnotesize \tt{y = \cancel \dfrac{171}{9} } \\ \\ \qquad \footnotesize \tt{y = 9} \\ \\ \footnotesize \tt{Putting \: y=9 \: in \: Equation (1)} \\ \\ \qquad \footnotesize \tt{3x - 5y = 6} \\ \\ \qquad \footnotesize \tt{3x - 5(9) = 6} \\ \\ \qquad \footnotesize \tt{3x - 45= 6} \\ \\ \qquad \footnotesize \tt{3x = 6 + 45} \\ \\ \qquad \footnotesize \tt{3x = 51} \\ \\ \qquad \footnotesize \tt{x = \cancel\dfrac{51}{3} } \\ \\ \qquad \footnotesize \tt{x = 17} \end{array}}}}\end{gathered}$