Explanation
Given equations are $$\dfrac { x }{ 2 } +\dfrac { y }{ 5 } =5$$ and $$ \dfrac { 2 }{ x } +\dfrac { 5 }{ y } =\dfrac { 5 }{ 6 } $$
Let $$\dfrac { x }{ 2 } =u$$ and $$\dfrac { y }{ 5 } =v$$
$$u+v=5$$ .....(i)
$$\dfrac { 1 }{ u } +\dfrac { 1 }{ v } =\dfrac { 5 }{ 6 }$$ .....(ii)
$$\Rightarrow \dfrac { u+v }{ uv } =\dfrac { 5 }{ 6 } \\ \Rightarrow \dfrac { 5 }{ uv } =\dfrac { 5 }{ 6 } \\ \Rightarrow uv=6$$ .........(iii)
From (i) $$v=5-u$$
$$(5-u)u=6\\ \Rightarrow 5u-{ u }^{ 2 }=6\\ \Rightarrow { u }^{ 2 }-5u+6=0\\ \Rightarrow { u }^{ 2 }-2u-3u+6=0\\ \Rightarrow u(u-2)-3(u-2)=0\\ \Rightarrow (u-3)(u-2)=0\\ \Rightarrow u=2,3$$
But $$ \dfrac { x }{ 2 } =u$$
$$\Rightarrow x=2u\\ \Rightarrow x=4,6$$
Also, $$ \dfrac { y }{ 5 } =v$$
$$ \Rightarrow y=5v\\ \Rightarrow y=15,10$$
$$let\>(\frac{1}{x+y})=u\>and\>(\frac{1}{x-y})=v\\then\>22u+165v=5\rightarrow\>(1.)\\and\>55+45v=14\rightarrow\>(2.)\\multiply\>equation\>(1.)by\>3,we\>get\\66u+45v=15\rightarrow\>(1.)\\+55u+45v=14\rightarrow\>(2.)\\upon\>subtraction\>of\>(1.)with(2.),we\>get\>\\\>11u+0=1\\\therefore\>u=(\frac{1}{11})or\>x+y\>=11\rightarrow\>(3.)\\and\>then\>\\v=\>(\frac{5-22u}{15})=(\frac{5-2}{15})=(\frac{1}{5})\\\therefore\>x-y=5\rightarrow\>(4.)\\add\>(3.)and\>(4.)equation\>we\>get\>\\2x=16\\\therefore\>x=8\\and\>y\>=11-x=11-8=3$$
$$\dfrac{x}{10}+\dfrac{y}{5}-1=0$$
$$\dfrac{x}{10}+\dfrac{y}{5}=1$$
$$\dfrac{x+2y}{10}=1$$
$$x+2y=10$$.....................(1)
now consider,$$\dfrac{x}{8}+\dfrac{y}{6}=15$$
$$\dfrac{6x+8y}{48}=15$$
$$3x+4y=24\times{15}=360$$.................................(2)
now multiply (1)by 3
$$3x+6y=30$$ ........................(3)
subtracting (2) and (3)
$$2y=-330$$
$$y=-165$$
put in(1)
$$x+2(-165)=10$$
$$x-330=10$$
$$x=340$$
now, $$y=\lambda{x}+5$$
$$\implies{-165=\lambda{(340)}+5}$$
$$-170=\lambda{(340)}$$
$$\lambda{=\dfrac{-170}{340}=\dfrac{-1}{2}}$$
$$3x+4y= 10 \ .....(1)$$
$$2x-2y=2\ \ .....(2)$$
equation $$(2)$$ can be written as
$$\Rightarrow x-y=1$$
$$\Rightarrow x=y+1\ .....(3)$$
substitute $$x$$ value in equation $$(2)$$
$$\Rightarrow 3x+4y= 10 $$
$$\Rightarrow 3(y+1)+4y=10$$
$$\Rightarrow 3y+3+4y=10$$
$$\Rightarrow 7y=7$$
$$\Rightarrow y=1$$
then, substitute $$y$$ value in equation $$(3)$$
$$\Rightarrow x=y+1=1+1=2$$
$$\therefore x=2, y=1$$
Plot each line by taking two points on them.$$2x-3y=1$$$$x=0\implies y =\frac{-1}{3}$$ and $$y=0\implies x=\frac{1}{2}$$Join the two points $$(0,\frac{-1}{3})$$ and $$(\frac{1}{2},0)$$ by drawing line.
$$4x-3y+1=0$$
$$x=0\implies y =\frac{-1}{3}$$and $$y=0\implies x=\frac{1}{4}$$
Join the two points $$(0,\frac{-1}{3})$$ and $$(\frac{1}{4},0)$$ by drawing line.
From the graph, the point of intersection of both lines is $$ (-1,-1)$$ and hence solution of given system of linear equation is $$x=-1, y=-1$$
$$\textbf{Step -1: Form the required equations.}$$
$$\text{Let the number of girls in class }A\text{ be }x.$$
$$\text{Let the number of girls in class }B\text{ be }y.$$
$$\text{Let the number of boys in class }A\text{ be }35-x.$$
$$\text{Let the number of boys in class }B\text{ be }35-y.$$
$$\text{If seven girls are shifted from class }A\text{ to }B,$$
$$\text{Number of girls in class }A\text{ would become }x-7.$$
$$\text{Number of girls in class }B\text{ would become }y+7.$$
$$\text{But according to the question,}$$
$$y=x-7$$
$$\Rightarrow x-y=7\ldots(i)$$
$$\text{If four girls are shifted from class }B\text{ to }A,$$
$$\text{Number of girls in class }A\text{ would become }x+4.$$
$$\text{Number of girls in class }B\text{ would become }y-4.$$
$$x+4=2y$$
$$\Rightarrow x-2y=-4\ldots(ii)$$
$$\textbf{Step -2: Solve the above formed two equations simultaneously.}$$
$$\text{On subtracting equation }(ii)\text{ from }(i),\text{ we get}$$
$$-y+2y=7+4$$
$$\Rightarrow y=11$$
$$\text{On putting the value of }y\text{ in equation }(i),\text{ we get}$$
$$x-11=7$$
$$\Rightarrow x=18$$
$$\textbf{Step -3: Find the number of boys in class A and B.}$$
$$\text{Number of boys in class }A=35-18$$
$$=17$$
$$\text{Number of boys in class }B=35-11$$
$$=24$$
$$\textbf{Therefore, option D is correct.}$$
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