CBSE Questions for Class 10 Maths Pair Of Linear Equations In Two Variables Quiz 16 - MCQExams.com

The given equations are
$$\Rightarrow \dfrac{16}{x+3}+\dfrac{3}{y-2}=5....eq\left ( 1 \right )$$
$$\Rightarrow \dfrac{8}{x+3}-\dfrac{1}{y-2}=0....eq\left ( 2\right )$$
$$Put   \dfrac{1}{x+3}=u   \  and   \ \ \dfrac{1}{y-2}=v \   in \   eq\left (1  \right )  and  \left ( 2\right )$$
$$\Rightarrow 16u+3v=5....eq\left ( 3 \right )$$
$$\Rightarrow 8u-v=0 ....eq\left ( 4\right )$$
$$multiply  \  eq\left (4  \right ) \  by   \ 3 \  and \   subtract\    eq  \left (3  \right )  and  \  eq\left ( 4 \right ) $$
$$\Rightarrow \left (16u+3v=5 \right )-\left ( 24u-v=0 \right )$$
$$\Rightarrow 40u=5\Rightarrow u=\dfrac{1}{8}$$
$$Put   u=\dfrac{1}{8} \    in  \  eq\left ( 3\right )$$
$$\Rightarrow 16\left (\dfrac{1}{8}  \right )+3v=5\Rightarrow 3v=3\Rightarrow v=1 $$
$$Hence ,\dfrac{1}{x+3}= u=\dfrac{1}{8}\Rightarrow x+3=8\Rightarrow x=5$$
$$      \dfrac{1}{y-2}= v=1\Rightarrow y-2=1\Rightarrow y=3$$ 
     

The given equations are
$$\Rightarrow \dfrac{24}{2x+y}-\dfrac{13}{3x+2y}=2...eq\left ( 1 \right )$$
$$\Rightarrow \dfrac{26}{3x+2y}+\dfrac{8}{2x+y}=3...eq\left ( 1 \right )$$
$$Put   \dfrac{1}{2x+y}=u \   and \   \dfrac{1}{3x+y}=v  \  in \   eq\left (1  \right )  and   \left ( 2 \right )$$
$$\Rightarrow 24u-13v=2.....\left ( 3 \right )$$
$$\Rightarrow 8u+26v=3......\left ( 4 \right )$$
$$Multiply  \  eq\left (  3\right ) \  by\    26  \  and   \  eq\left (4  \right ) \  by\   13   \ and\    subtract   \ both$$
$$\Rightarrow \left (624u-338v=52  \right )-\left ( 104u+338v=39 \right )$$
$$\Rightarrow  728u=91\Rightarrow u=\dfrac{1}{8}$$
$$Put u=\dfrac{1}{8}  \  in \   eq\left ( 4 \right )$$
$$\Rightarrow 8\times\dfrac{1}{8}+26v=3\Rightarrow 26v=2\Rightarrow v=\dfrac{1}{13}$$
$$Hence, u=\dfrac{1}{2x+y}=\dfrac{1}{8} \Rightarrow 2x+y=8.....eq\left ( 5 \right )$$
$$       v=\dfrac{1}{3x+2y}=\dfrac{1}{13}\Rightarrow 3x+2y=13...eq\left ( 6 \right ) $$
$$Multiply  \  eq\left (5  \right ) \  by  \  2 \   and  \  subtract \   from  \  eq\left ( 6 \right )$$
$$\Rightarrow \left ( 4x+2y=16 \right )-\left ( 3x+3y=13 \right )$$
$$\Rightarrow x=3$$
$$put  \  x=3  \  in \   eq\left (5  \right )$$
$$\Rightarrow 2\times 3+y=8 \Rightarrow y=2$$