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=1x=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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