Question

Solve the following pair of liner equation by the substitution method.
$\frac{3x}{2 } - \frac{5y}{3} = - 2 \\ \\ \frac{x}{3} + \frac{y}{2} = \frac{13}{6}$
Please give the answer fast its very urgent​

Answer 1

AnswEr :

$\underline{\bigstar\:\textsf{According to the Question :}}$

$\displaystyle:\implies\sf\frac{x}{3} + \frac{y}{2} = \frac{13}{6}\\\\\\\displaystyle:\implies\sf\frac{2x + 3y}{6} = \frac{13}{6}\\\\\\\displaystyle:\implies\sf2x + 3y = 13\\\\\\\displaystyle:\implies\sf2x = 13 - 3y\\\\\\\displaystyle:\implies\sf x = \frac{13 - 3y}{2}$

$\rule{200}{1}$

$\underline{\bigstar\:\textsf{Putting the value of x in another equation :}}$

$\displaystyle:\implies\sf \frac{3x}{2 } - \frac{5y}{3} = - 2\\\\\\\displaystyle:\implies\sf \frac{3(13 - 3y)}{2 \times 2} - \frac{5y}{3} = - 2\\\\\\\displaystyle:\implies\sf \frac{39 - 9y}{4} - \frac{5y}{3} = - 2\\\\\\\displaystyle:\implies\sf \frac{3(39 - 9y) - 4(5y)}{12} = - 2\\\\\\\displaystyle:\implies\sf 117 - 27y - 20y = - 2 \times 12\\\\\\\displaystyle:\implies\sf117 - 47y = - 24\\\\\\\displaystyle:\implies\sf - 47y = - 24 - 117\\\\\\\displaystyle:\implies\sf - 47y = - 141\\\\\\\displaystyle:\implies\sf y = \frac{ - 141}{ - 47}\\\\\\\displaystyle:\implies\boxed{\pink{\sf y = 3}}$

$\rule{150}{2}$

$\underline{\bigstar\:\textsf{Value of x :}}$

$\displaystyle:\implies\sf x = \frac{13 -3y}{2}\\\\\\\displaystyle:\implies\sf x = \frac{13 - 3(3)}{2}\\\\\\\displaystyle:\implies\sf x = \frac{13 - 9}{2}\\\\\\\displaystyle:\implies\sf x = \frac{4}{2}\\\\\\\displaystyle:\implies\boxed{\pink{\sf x = 2}}$

⠀

$\therefore\:\underline{\textsf{Value of x and y is \textbf{2 and 3 respectively.}}}$

Answer 2

$\bf{\Huge{\boxed{\tt{\purple{ANSWER\::}}}}}$

$\bf{\Large{\underline{\bf{Given\::}}}}$

$\sf{\dfrac{3x}{2} \:-\:\dfrac{5y}{3} \:=\:-2\:\:\:\:\:\:\:\:\&\:\:\:\:\:\:\:\dfrac{x}{3} +\dfrac{y}{2}=\dfrac{13}{6} }$

$\bf{\Large{\underline{\bf{To\:find\::}}}}$

The substitution Method.

$\bf{\Large{\underline{\tt{\red{Explanation\::}}}}}$

We have,

$\sf{\dfrac{3x}{2} \:-\:\dfrac{5y}{3} \:=\:-2.........................(1)}\\\\\\\sf{\dfrac{x}{3} \:+\:\dfrac{y}{2} \:=\:\dfrac{13}{6}.............................(2) }$

From Equation (1), we get;

$\mapsto\:\sf{\dfrac{3x}{2} -\dfrac{5y}{3}=-2}\\\\\\\mapsto\:\sf{\dfrac{9x-10y}{6} =-2}\\\\\\\mapsto\:\sf{9x-10y=-12}\\\\\\\mapsto\:\sf{9x=-12+10y}\\\\\\\mapsto\:\sf{\red{x\:=\:\dfrac{-12+10y}{9}..................(3) }}$

Putting the value of x in equation (2), we get;

$\mapsto\:\sf{\dfrac{\dfrac{-12+10y}{9} }{3}\: +\:\dfrac{y}{2} =\dfrac{13}{6} }$

$\mapsto\:\sf{\dfrac{-12+10y}{27} +\dfrac{y}{2} =\dfrac{13}{6} }$

$\mapsto\:\sf{\dfrac{-24+20y+27y}{54} =\dfrac{13}{6} }$

$\mapsto\:\sf{\dfrac{-24+47y}{54} =\dfrac{13}{6} }$

$\bf{\Large{\boxed{\sf{\orange{\bigstar{Cross\:-\:Multiplication\::}}}}}}}$

$\mapsto\:\sf{6(-24+47y)\:=\:(13*54)}$

$\mapsto\:\sf{-144+282y\:=\:702}$

$\mapsto\:\sf{282y\:=\:702\:+\:144}$

$\mapsto\:\sf{282y\:=\:846}$

$\mapsto\:\sf{y\:=\:\cancel{\dfrac{846}{282} }}$

$\mapsto\:\sf{\red{y\:=\:3}}$

Putting the value of y in equation (3), we get;

$\longmapsto\sf{x\:=\:\dfrac{-12+10(3)}{9} }\\\\\\\longmapsto\sf{x\:=\:\dfrac{-12+30}{9} }\\\\\\\\\longmapsto\sf{x\:=\:\cancel{\dfrac{18}{9} }}\\\\\\\\\longmapsto\sf{\red{x\:=\:2}}$

Thus,

The value is x = 2 & y = 3.