Pair of Linear Equations in Two Variables Test - 41

The given equations are
$$\Rightarrow 6x+5y=8xy....eq\left (1  \right )$$

$$\Rightarrow 8x+3y=7xy....eq\left (2  \right )$$

$$Divide \   both \   equations  \  by \ \     xy  $$

$$\Rightarrow \dfrac{6}{y}+\dfrac{5}{x}=8....eq\left ( 3 \right )$$

$$\Rightarrow \dfrac{8}{y}+\dfrac{3}{1x}=7....eq\left ( 4 \right )$$

$$Put\ \dfrac{1}{x}=u\  and\ \dfrac{1}{y}=v\  in\   eq\left (3  \right )  and  \left ( 4 \right )$$

$$\Rightarrow 6v+5u=8....eq\left ( 5 \right )$$

$$\Rightarrow 8v+3u=7 ....eq\left ( 6\right )$$

$$Multiply \   eq \   \left (5  \right )\   by   \  3  \  and  \  eq\left ( 6 \right )  \  by\    5 \   and  \  subtract \   both$$

$$\Rightarrow \left (18v+15u=24 \right )-\left ( 40v+15u=35 \right )$$

$$\Rightarrow -22v=-11\Rightarrow v=\dfrac{1}{2}$$

$$Put  \  v=\dfrac{1}{2}  \  in  \  eq\left ( 5\right )$$

$$\Rightarrow 6\left ( \dfrac{1}{2} \right )+5u=8\Rightarrow u=1$$

$$Hence,\dfrac{1}{x}= u=1\Rightarrow x=1$$

$$      \dfrac{1}{y}= v=\dfrac{1}{2}\Rightarrow y=2$$
     

The given equations are
$$\Rightarrow \dfrac{11}{2x}-\dfrac{9}{2y}=-\dfrac{23}{2}....eq\left ( 1 \right )$$

$$\Rightarrow \dfrac{3}{4x}+\dfrac{7}{15y}=-\dfrac{23}{6}....eq\left ( 2 \right )$$

Put   $$u=\dfrac{1}{x} $$  and  $$ v=\dfrac{1}{y} $$ in  $$ eq\left (1  \right )$$  and $$ \left ( 2 \right )$$

$$\Rightarrow \dfrac{11}{2}u-\dfrac{9}{2}v=-\dfrac{23}{2}\Rightarrow 11u-9v=-23....eq\left ( 3 \right )$$

$$\Rightarrow \dfrac{3}{4}u+\dfrac{7}{15}v=\dfrac{23}{6}\Rightarrow 45u+28v=230 ....eq\left ( 4 \right )$$

Multiply   $$eq \left (3  \right )$$  by    28   and   $$eq\left ( 4 \right )$$   by   9   and   subtract   both

$$\Rightarrow \left ( 308u-252v=-644 \right )-\left ( 405u+252v=2070 \right )$$

$$\Rightarrow 713u=1426\Rightarrow u=2$$

$$Put\   u=2\   in\   eq\left ( 3 \right )$$

$$\Rightarrow 11\times2-9v=-23\Rightarrow v=5$$

$$Hence,  \dfrac{1}{x}= u=2\Rightarrow x=\dfrac{1}{2}$$
    $$  \dfrac{1}{y}= v=5\Rightarrow y=\dfrac{1}{5}$$
     

Given systems of equations are:
$$9+25xy = 53x$$
$$27-4xy = x$$
Now using reducible method will get
Let convert this equation in $$\dfrac{1}{x}\  and\  \dfrac{1}{y}$$ form
Will take $$xy$$ as $$L.C.M$$ as $$xy$$ is common
From eq  will get 
$$\Rightarrow  \dfrac{9}{xy} +25 = \dfrac{53}{y}$$ ...3
$$\dfrac{27}{xy} - 4 = \dfrac{1}{y}$$  ...4
Now let  $$\dfrac{1}{xy}  = v  and  \dfrac{1}{y} = u $$
Now will get 
$$9v+25 = 53u $$  and  $$27v-4 = u$$
Now Rearranging both the equation
$$53u-9v = 25 ...(5)$$
$$u-27v = -4 ....(6)$$
Now multiplying eq5 by 3 
$$\Rightarrow 159u-27v = 75  ...(7)$$
Now  subtracting eq 7 and eq6
$$\Rightarrow 159u-27v = 75$$
$$\Rightarrow -u +27v = 4$$
$$\Rightarrow 158u = 79$$
$$\Rightarrow u = \dfrac{1}{2}$$
Now putting $$u = \dfrac{1}{2}$$ in eq 6
$$\Rightarrow  \dfrac{1}{2} -27\left ( 2 \right ) v= -4\left ( 2 \right )$$
$$\Rightarrow  1 - 54v = -8$$
$$\Rightarrow  54v =  9$$
$$\Rightarrow  v = \dfrac{9}{54}$$
$$\Rightarrow v  = \dfrac{1}{6}$$
Now putting u and v values
$$\dfrac{1}{y} = u$$
$$ \Rightarrow \dfrac{1}{y} = \dfrac{1}{2}$$
$$\Rightarrow y = 2$$
$$\dfrac{1}{xy} = v$$
$$\Rightarrow \dfrac{1}{xy} = \dfrac{1}{6}$$
$$\Rightarrow \dfrac{1}{2x} = \dfrac{1}{6}$$
$$ \Rightarrow x = \dfrac{6}{2}$$
$$\Rightarrow x = 3$$
$$\Rightarrow x = 3 , y = 2$$

The given equations are
$$\Rightarrow \frac{2}{x}+\frac{3}{y}=2....eq\left ( 1 \right )$$
$$\Rightarrow  \frac{1}{x}-\frac{1}{2y}=\frac{1}{3}....eq\left ( 2 \right )$$
$$Put   \frac{1}{x}=u   \ and \   \frac{1}{y}=v   \ in\    eq\left ( 1 \right ) \ and  \ \left (2  \right )$$
$$\Rightarrow 2u+2v=2....eq\left (3  \right )$$
$$\Rightarrow u-\frac{1}{2}v=\frac{1}{3}\Rightarrow 6u-3v=2.....eq\left (4  \right )$$
$$Substract   \ eq\left ( 3 \right ) \   and \   eq\left ( 4 \right )$$
$$\Rightarrow 8u=4\Rightarrow u=\frac{1}{2}$$
$$put  \  the\    value \   of  \  u  \  in  \  eq\left ( 3 \right )$$
$$\Rightarrow 2\times \frac{1}{2}+3v=2\Rightarrow v =\frac{1}{3}$$
$$Hence, \frac{1}{x}=u=\frac{1}{2}\Rightarrow x=2$$
$$       \frac{1}{y}=v=\frac{1}{2}\Rightarrow y=3$$

$$2y-x=8$$

$$4x-3y+4=0$$

The given equations are
$$\Rightarrow \dfrac{2}{x-1}+\dfrac{y-2}{4}=2......eq(1)$$
$$\Rightarrow \dfrac{3}{2\left (x-1  \right )}\dfrac{2\left (y-2  \right )}{4}=\dfrac{47}{20}......eq(2)$$
$$Put   \dfrac{1}{x-1}=u   \ and  \  y-2=v \   in \   eq(1)\    and \   eq(2)$$
$$\Rightarrow 2u+\dfrac{v}{4}=2\Rightarrow 8u+v=8....eq(3)$$
$$\Rightarrow \dfrac{3u}{2}+\dfrac{2v}{5}=\dfrac{47}{20}\Rightarrow 30u+8v=47.....eq(4)$$
$$Multiply  \  eq(3)  \  by \   8$$
$$\Rightarrow 64u+8v=64.....eq(5)$$
$$Subtract  \  eq(4) \   and \   eq(5)$$
$$\Rightarrow \left (64u+8v=64  \right )-\left (30u+8v+47  \right )$$
$$\Rightarrow 34u=17\Rightarrow u=\dfrac{1}{2}$$
$$put  \  u=\dfrac{1}{2}  \  in \   eq(3)$$
$$\Rightarrow 8\times \dfrac{1}{2}+v=8\Rightarrow v=4$$
$$Hence, u=\dfrac{1}{x-1} =\dfrac{1}{2}\Rightarrow x-1=2\Rightarrow x=3$$
$$       v=y-2=4\Rightarrow y=6  $$
  
 

The given equations are
$$\Rightarrow \dfrac{x-y}{xy}=9\Rightarrow x-y=9xy....eq\left (1  \right )$$
$$\Rightarrow \dfrac{x+y}{xy}=5\Rightarrow x+y=5xy....eq\left (2  \right )$$
Divide both eq by $$xy$$
$$\Rightarrow \displaystyle \frac{1}{y}-\frac{1}{x}=9....eq\left ( 3 \right )$$
$$\Rightarrow \displaystyle \frac{1}{y}+\frac{1}{x}=5....eq\left ( 4 \right )$$
Put $$\dfrac{1}{x}=u$$ and $$\dfrac{1}{y}=v$$ in eq $$\left (3  \right )$$ and $$\left ( 4 \right )$$
$$\Rightarrow v-u=9 ....eq\left ( 5 \right )$$
$$\Rightarrow v+u=5 ....eq\left ( 6\right )$$
Subtract eq $$\left (5  \right )$$ and eq$$\left ( 6 \right )$$
$$\Rightarrow \left (v-u=9 \right )-\left ( v+u=5 \right )$$
$$\Rightarrow 2v=-14\Rightarrow v=7$$
Put $$v=7$$ in eq$$\left ( 5\right )$$
$$\Rightarrow 7+u=5\Rightarrow u=-2$$
Hence, $$\dfrac{1}{x}= u=-2\Rightarrow x=-\dfrac{1}{2}$$
$$      \dfrac{1}{y}= v=7\Rightarrow y=\dfrac{1}{7}$$
     

The given equations are
$$\Rightarrow \dfrac{1}{3x}-\dfrac{1}{7y}=\dfrac{2}{3}....eq\left ( 1 \right )$$
$$\Rightarrow \dfrac{1}{2x}-\dfrac{1}{3y}=\dfrac{1}{6}....eq\left (2  \right )$$
Put   $$\dfrac{1}{x}=u$$   and   $$\dfrac{1}{y}=v$$   in   $$eq\left ( 1 \right ) $$  and  $$eq \left (2  \right )$$
$$\Rightarrow  \dfrac{1}{3}u-\dfrac{1}{7}v=\dfrac{2}{3}\Rightarrow 7u-3v=14....eq\left ( 3 \right )$$
$$\Rightarrow \dfrac{1}{2}u-\dfrac{1}{3}v=\dfrac{1}{6}\Rightarrow 3u-2v=1....eq\left (4  \right )$$
$$Substract\   eq\left ( 3 \right )   and\   eq\left ( 4 \right )$$
$$\Rightarrow 5u=25\Rightarrow u=5$$
$$put\   the\   value\   of\  u\   in\   eq\left ( 3 \right )$$
$$\Rightarrow 7\times 5-3v=14\Rightarrow v =7$$
$$Hence, \dfrac{1}{x}=u=5\Rightarrow x=\dfrac{1}{5}$$
$$       \dfrac{1}{y}=v=7\Rightarrow y=\dfrac{1}{7}$$

x=3,y=2.