\displaystyle \textbf{Question 1: }\text{Find the values of }x\text{ and }y\text{ in the following rectangle.} 
\displaystyle \text{Answer:}
\displaystyle \text{Since the opposite sides of a rectangle are equal,}
\displaystyle AB=DC
\displaystyle \Rightarrow x+3y=13\qquad\ldots\text{(i)}
\displaystyle AD=BC
\displaystyle \Rightarrow 3x+y=7\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 3x+9y=39\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle 8y=32
\displaystyle \Rightarrow y=4
\displaystyle \text{Substituting }y=4\text{ in equation (ii),}
\displaystyle 3x+4=7
\displaystyle \Rightarrow 3x=3
\displaystyle \Rightarrow x=1
\displaystyle \therefore x=1,\qquad y=4.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Solve the following system of equations:}
\displaystyle 11x+15y+23=0,\qquad 7x-2y-20=0
\displaystyle \text{Answer:}
\displaystyle \text{The given equations are}
\displaystyle 11x+15y=-23\qquad\ldots\text{(i)}
\displaystyle 7x-2y=20\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 22x+30y=-46\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }15,
\displaystyle 105x-30y=300\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 127x=254
\displaystyle \Rightarrow x=2
\displaystyle \text{Substituting }x=2\text{ in equation (ii),}
\displaystyle 14-2y=20
\displaystyle \Rightarrow -2y=6
\displaystyle \Rightarrow y=-3
\displaystyle \therefore x=2,\qquad y=-3.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Solve the following system of equations:}
\displaystyle \frac{x}{2}+y=0.8,\qquad\frac{7}{x+\frac{y}{2}}=10
\displaystyle \text{Answer:}
\displaystyle \frac{x}{2}+y=0.8
\displaystyle \Rightarrow 5x+10y=8\qquad\ldots\text{(i)}
\displaystyle \frac{7}{x+\frac{y}{2}}=10
\displaystyle \Rightarrow 7=10\left(x+\frac{y}{2}\right)
\displaystyle \Rightarrow 10x+5y=7\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 10x+20y=16\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle 15y=9
\displaystyle \Rightarrow y=\frac{3}{5}
\displaystyle \text{Substituting }y=\frac{3}{5}\text{ in equation (ii),}
\displaystyle 10x+5\left(\frac{3}{5}\right)=7
\displaystyle \Rightarrow 10x+3=7
\displaystyle \Rightarrow 10x=4
\displaystyle \Rightarrow x=\frac{2}{5}
\displaystyle \therefore x=\frac{2}{5},\qquad y=\frac{3}{5}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Solve the following system of equations:}
\displaystyle \frac{x}{3}+\frac{y}{4}=11,\qquad\frac{5x}{6}-\frac{y}{3}=-7
\displaystyle \text{Answer:}
\displaystyle \frac{x}{3}+\frac{y}{4}=11
\displaystyle \Rightarrow 4x+3y=132\qquad\ldots\text{(i)}
\displaystyle \frac{5x}{6}-\frac{y}{3}=-7
\displaystyle \Rightarrow 5x-2y=-42\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 8x+6y=264\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,
\displaystyle 15x-6y=-126\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 23x=138
\displaystyle \Rightarrow x=6
\displaystyle \text{Substituting }x=6\text{ in equation (i),}
\displaystyle 4(6)+3y=132
\displaystyle \Rightarrow 3y=108
\displaystyle \Rightarrow y=36
\displaystyle \therefore x=6,\qquad y=36.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Solve the following system of equations:}
\displaystyle \frac{4}{x}+3y=8,\qquad\frac{6}{x}-4y=-5\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x}=u.
\displaystyle \text{The given equations become}
\displaystyle 4u+3y=8\qquad\ldots\text{(i)}
\displaystyle 6u-4y=-5\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }4,
\displaystyle 16u+12y=32\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,
\displaystyle 18u-12y=-15\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 34u=17
\displaystyle \Rightarrow u=\frac{1}{2}
\displaystyle \therefore \frac{1}{x}=\frac{1}{2}\Rightarrow x=2
\displaystyle \text{Substituting }u=\frac{1}{2}\text{ in equation (i),}
\displaystyle 4\left(\frac{1}{2}\right)+3y=8
\displaystyle \Rightarrow 2+3y=8
\displaystyle \Rightarrow y=2
\displaystyle \therefore x=2,\qquad y=2.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Solve the following system of equations:}
\displaystyle 3x-\frac{y+7}{11}+2=10,\qquad2y+\frac{x+11}{7}=10
\displaystyle \text{Answer:}
\displaystyle 3x-\frac{y+7}{11}+2=10
\displaystyle \Rightarrow 33x-y-7=88
\displaystyle \Rightarrow 33x-y=95\qquad\ldots\text{(i)}
\displaystyle 2y+\frac{x+11}{7}=10
\displaystyle \Rightarrow 14y+x+11=70
\displaystyle \Rightarrow x+14y=59\qquad\ldots\text{(ii)}
\displaystyle \text{From equation (i),}
\displaystyle y=33x-95
\displaystyle \text{Substituting }y=33x-95\text{ in equation (ii),}
\displaystyle x+14(33x-95)=59
\displaystyle \Rightarrow x+462x-1330=59
\displaystyle \Rightarrow 463x=1389
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (i),}
\displaystyle 33(3)-y=95
\displaystyle \Rightarrow 99-y=95
\displaystyle \Rightarrow y=4
\displaystyle \therefore x=3,\qquad y=4.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Solve the following system of equations:}
\displaystyle \frac{4}{x}+3y=14,\qquad\frac{3}{x}-4y=23 
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x}=u.
\displaystyle \text{The given equations become}
\displaystyle 4u+3y=14\qquad\ldots\text{(i)}
\displaystyle 3u-4y=23\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }4,
\displaystyle 16u+12y=56\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,
\displaystyle 9u-12y=69\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 25u=125
\displaystyle \Rightarrow u=5
\displaystyle \therefore \frac{1}{x}=5\Rightarrow x=\frac{1}{5}
\displaystyle \text{Substituting }u=5\text{ in equation (i),}
\displaystyle 4(5)+3y=14
\displaystyle \Rightarrow 20+3y=14
\displaystyle \Rightarrow 3y=-6
\displaystyle \Rightarrow y=-2
\displaystyle \therefore x=\frac{1}{5},\qquad y=-2.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Solve the following system of equations:}
\displaystyle \frac{2}{x}+\frac{3}{y}=13,\qquad\frac{5}{x}-\frac{4}{y}=-2 
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{The given equations become}
\displaystyle 2u+3v=13\qquad\ldots\text{(i)}
\displaystyle 5u-4v=-2\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }4,
\displaystyle 8u+12v=52\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,
\displaystyle 15u-12v=-6\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 23u=46
\displaystyle \Rightarrow u=2
\displaystyle \text{Substituting }u=2\text{ in equation (i),}
\displaystyle 4+3v=13
\displaystyle \Rightarrow 3v=9
\displaystyle \Rightarrow v=3
\displaystyle \therefore \frac{1}{x}=2\Rightarrow x=\frac{1}{2}
\displaystyle \text{and }\frac{1}{y}=3\Rightarrow y=\frac{1}{3}.
\displaystyle \therefore x=\frac{1}{2},\qquad y=\frac{1}{3}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Solve the following system of equations:}
\displaystyle \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2,\qquad\frac{4}{\sqrt{x}}-\frac{9}{\sqrt{y}}=-1 
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{\sqrt{x}}=u\text{ and }\frac{1}{\sqrt{y}}=v.
\displaystyle \text{The given equations become}
\displaystyle 2u+3v=2\qquad\ldots\text{(i)}
\displaystyle 4u-9v=-1\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 4u+6v=4\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (iii),}
\displaystyle 15v=5
\displaystyle \Rightarrow v=\frac{1}{3}
\displaystyle \text{Substituting }v=\frac{1}{3}\text{ in equation (i),}
\displaystyle 2u+3\left(\frac{1}{3}\right)=2
\displaystyle \Rightarrow 2u=1
\displaystyle \Rightarrow u=\frac{1}{2}
\displaystyle \therefore \frac{1}{\sqrt{x}}=\frac{1}{2}\Rightarrow \sqrt{x}=2\Rightarrow x=4
\displaystyle \text{and }\frac{1}{\sqrt{y}}=\frac{1}{3}\Rightarrow \sqrt{y}=3\Rightarrow y=9.
\displaystyle \therefore x=4,\qquad y=9.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Solve the following system of equations:}
\displaystyle 0.5x+0.7y=0.74,\qquad0.3x+0.5y=0.5
\displaystyle \text{Answer:}
\displaystyle 0.5x+0.7y=0.74
\displaystyle \Rightarrow 50x+70y=74
\displaystyle \Rightarrow 25x+35y=37\qquad\ldots\text{(i)}
\displaystyle 0.3x+0.5y=0.5
\displaystyle \Rightarrow 3x+5y=5\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (ii) by }7,
\displaystyle 21x+35y=35\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (i),}
\displaystyle 4x=2
\displaystyle \Rightarrow x=\frac{1}{2}=0.5
\displaystyle \text{Substituting }x=\frac{1}{2}\text{ in equation (ii),}
\displaystyle 3\left(\frac{1}{2}\right)+5y=5
\displaystyle \Rightarrow 5y=\frac{7}{2}
\displaystyle \Rightarrow y=\frac{7}{10}=0.7
\displaystyle \therefore x=0.5,\qquad y=0.7.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Solve the following system of equations:}
\displaystyle \sqrt{2}x-\sqrt{3}y=0,\qquad\sqrt{3}x-\sqrt{8}y=0
\displaystyle \text{Answer:}
\displaystyle \sqrt{2}x-\sqrt{3}y=0\qquad\ldots\text{(i)}
\displaystyle \sqrt{3}x-\sqrt{8}y=0\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }\sqrt{3},
\displaystyle \sqrt{6}x-3y=0\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }\sqrt{2},
\displaystyle \sqrt{6}x-4y=0\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle y=0
\displaystyle \text{Substituting }y=0\text{ in equation (i),}
\displaystyle \sqrt{2}x=0
\displaystyle \Rightarrow x=0
\displaystyle \therefore x=0,\qquad y=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Solve the following system of equations:}
\displaystyle \frac{xy}{x+y}=\frac{6}{5},\qquad\frac{xy}{y-x}=6
\displaystyle \text{Answer:}
\displaystyle \frac{xy}{x+y}=\frac{6}{5}
\displaystyle \Rightarrow \frac{x+y}{xy}=\frac{5}{6}
\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{5}{6}\qquad\ldots\text{(i)}
\displaystyle \frac{xy}{y-x}=6
\displaystyle \Rightarrow \frac{y-x}{xy}=\frac{1}{6}
\displaystyle \Rightarrow \frac{1}{x}-\frac{1}{y}=\frac{1}{6}\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Equations (i) and (ii) become}
\displaystyle u+v=\frac{5}{6}\qquad\ldots\text{(iii)}
\displaystyle u-v=\frac{1}{6}\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2u=1
\displaystyle \Rightarrow u=\frac{1}{2}
\displaystyle \text{Substituting }u=\frac{1}{2}\text{ in equation (iii),}
\displaystyle \frac{1}{2}+v=\frac{5}{6}
\displaystyle \Rightarrow v=\frac{1}{3}
\displaystyle \therefore \frac{1}{x}=\frac{1}{2}\Rightarrow x=2
\displaystyle \text{and }\frac{1}{y}=\frac{1}{3}\Rightarrow y=3.
\displaystyle \therefore x=2,\qquad y=3.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Solve the following system of equations:}
\displaystyle x+y=2xy,\qquad\frac{x-y}{xy}=6
\displaystyle \text{Answer:}
\displaystyle x+y=2xy
\displaystyle \Rightarrow \frac{x+y}{xy}=2
\displaystyle \Rightarrow \frac{1}{x}+\frac{1}{y}=2\qquad\ldots\text{(i)}
\displaystyle \frac{x-y}{xy}=6
\displaystyle \Rightarrow \frac{1}{y}-\frac{1}{x}=6\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Equations (i) and (ii) become}
\displaystyle u+v=2\qquad\ldots\text{(iii)}
\displaystyle v-u=6\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2v=8
\displaystyle \Rightarrow v=4
\displaystyle \text{Substituting }v=4\text{ in equation (iii),}
\displaystyle u+4=2
\displaystyle \Rightarrow u=-2
\displaystyle \therefore \frac{1}{x}=-2\Rightarrow x=-\frac{1}{2}
\displaystyle \text{and }\frac{1}{y}=4\Rightarrow y=\frac{1}{4}.
\displaystyle \therefore x=-\frac{1}{2},\qquad y=\frac{1}{4}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{Solve the following system of equations:}
\displaystyle \frac{44}{x+y}+\frac{30}{x-y}=10,\qquad\frac{55}{x+y}+\frac{40}{x-y}=13\hfill\text{[CBSE 2002C]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x+y}=u\text{ and }\frac{1}{x-y}=v.
\displaystyle \text{The given equations become}
\displaystyle 44u+30v=10
\displaystyle \Rightarrow 22u+15v=5\qquad\ldots\text{(i)}
\displaystyle 55u+40v=13\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }5,
\displaystyle 110u+75v=25\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }2,
\displaystyle 110u+80v=26\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iii) from equation (iv),}
\displaystyle 5v=1
\displaystyle \Rightarrow v=\frac{1}{5}
\displaystyle \text{Substituting }v=\frac{1}{5}\text{ in equation (i),}
\displaystyle 22u+15\left(\frac{1}{5}\right)=5
\displaystyle \Rightarrow 22u+3=5
\displaystyle \Rightarrow u=\frac{1}{11}
\displaystyle \therefore \frac{1}{x+y}=\frac{1}{11}\Rightarrow x+y=11\qquad\ldots\text{(v)}
\displaystyle \frac{1}{x-y}=\frac{1}{5}\Rightarrow x-y=5\qquad\ldots\text{(vi)}
\displaystyle \text{Adding equations (v) and (vi),}
\displaystyle 2x=16
\displaystyle \Rightarrow x=8
\displaystyle \text{Substituting }x=8\text{ in equation (v),}
\displaystyle 8+y=11
\displaystyle \Rightarrow y=3
\displaystyle \therefore x=8,\qquad y=3.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Solve the following system of equations:}
\displaystyle \frac{5}{x-1}+\frac{1}{y-2}=2,\qquad\frac{6}{x-1}-\frac{3}{y-2}=1\hfill\text{[CBSE 2009]}
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x-1}=u\text{ and }\frac{1}{y-2}=v.
\displaystyle \text{The given equations become}
\displaystyle 5u+v=2\qquad\ldots\text{(i)}
\displaystyle 6u-3v=1\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }3,
\displaystyle 15u+3v=6\qquad\ldots\text{(iii)}
\displaystyle \text{Adding equations (ii) and (iii),}
\displaystyle 21u=7
\displaystyle \Rightarrow u=\frac{1}{3}
\displaystyle \text{Substituting }u=\frac{1}{3}\text{ in equation (i),}
\displaystyle \frac{5}{3}+v=2
\displaystyle \Rightarrow v=\frac{1}{3}
\displaystyle \therefore \frac{1}{x-1}=\frac{1}{3}\Rightarrow x-1=3\Rightarrow x=4
\displaystyle \text{and }\frac{1}{y-2}=\frac{1}{3}\Rightarrow y-2=3\Rightarrow y=5.
\displaystyle \therefore x=4,\qquad y=5.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{Solve the following system of equations:}
\displaystyle \frac{10}{x+y}+\frac{2}{x-y}=4,\qquad\frac{15}{x+y}-\frac{9}{x-y}=-2 
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{x+y}=u\text{ and }\frac{1}{x-y}=v.
\displaystyle \text{The given equations become}
\displaystyle 10u+2v=4
\displaystyle \Rightarrow 5u+v=2\qquad\ldots\text{(i)}
\displaystyle 15u-9v=-2\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }9,
\displaystyle 45u+9v=18\qquad\ldots\text{(iii)}
\displaystyle \text{Adding equations (ii) and (iii),}
\displaystyle 60u=16
\displaystyle \Rightarrow u=\frac{4}{15}
\displaystyle \text{Substituting }u=\frac{4}{15}\text{ in equation (i),}
\displaystyle 5\left(\frac{4}{15}\right)+v=2
\displaystyle \Rightarrow \frac{4}{3}+v=2
\displaystyle \Rightarrow v=\frac{2}{3}
\displaystyle \therefore \frac{1}{x+y}=\frac{4}{15}\Rightarrow x+y=\frac{15}{4}\qquad\ldots\text{(iv)}
\displaystyle \frac{1}{x-y}=\frac{2}{3}\Rightarrow x-y=\frac{3}{2}\qquad\ldots\text{(v)}
\displaystyle \text{Adding equations (iv) and (v),}
\displaystyle 2x=\frac{15}{4}+\frac{3}{2}=\frac{21}{4}
\displaystyle \Rightarrow x=\frac{21}{8}
\displaystyle \text{Subtracting equation (v) from equation (iv),}
\displaystyle 2y=\frac{15}{4}-\frac{3}{2}=\frac{9}{4}
\displaystyle \Rightarrow y=\frac{9}{8}
\displaystyle \therefore x=\frac{21}{8},\qquad y=\frac{9}{8}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Solve the following system of equations:}
\displaystyle \frac{1}{3x+y}+\frac{1}{3x-y}=\frac{3}{4},\qquad\frac{1}{2(3x+y)}-\frac{1}{2(3x-y)}=-\frac{1}{8} 
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{1}{3x+y}=u\text{ and }\frac{1}{3x-y}=v.
\displaystyle \text{The given equations become}
\displaystyle u+v=\frac{3}{4}\qquad\ldots\text{(i)}
\displaystyle \frac{u}{2}-\frac{v}{2}=-\frac{1}{8}
\displaystyle \Rightarrow u-v=-\frac{1}{4}\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 2u=\frac{1}{2}
\displaystyle \Rightarrow u=\frac{1}{4}
\displaystyle \text{Substituting }u=\frac{1}{4}\text{ in equation (i),}
\displaystyle \frac{1}{4}+v=\frac{3}{4}
\displaystyle \Rightarrow v=\frac{1}{2}
\displaystyle \therefore \frac{1}{3x+y}=\frac{1}{4}\Rightarrow3x+y=4\qquad\ldots\text{(iii)}
\displaystyle \frac{1}{3x-y}=\frac{1}{2}\Rightarrow3x-y=2\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 6x=6
\displaystyle \Rightarrow x=1
\displaystyle \text{Substituting }x=1\text{ in equation (iii),}
\displaystyle 3+y=4
\displaystyle \Rightarrow y=1
\displaystyle \therefore x=1,\qquad y=1.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Solve the following system of equations:}
\displaystyle \frac{7x-2y}{xy}=5,\qquad\frac{8x+7y}{xy}=15 
\displaystyle \text{Answer:}
\displaystyle \frac{7x-2y}{xy}=5
\displaystyle \Rightarrow \frac{7}{y}-\frac{2}{x}=5\qquad\ldots\text{(i)}
\displaystyle \frac{8x+7y}{xy}=15
\displaystyle \Rightarrow \frac{8}{y}+\frac{7}{x}=15\qquad\ldots\text{(ii)}
\displaystyle \text{Let }\frac{1}{x}=u\text{ and }\frac{1}{y}=v.
\displaystyle \text{Equations (i) and (ii) become}
\displaystyle -2u+7v=5\qquad\ldots\text{(iii)}
\displaystyle 7u+8v=15\qquad\ldots\text{(iv)}
\displaystyle \text{Multiplying equation (iii) by }7,
\displaystyle -14u+49v=35\qquad\ldots\text{(v)}
\displaystyle \text{Multiplying equation (iv) by }2,
\displaystyle 14u+16v=30\qquad\ldots\text{(vi)}
\displaystyle \text{Adding equations (v) and (vi),}
\displaystyle 65v=65
\displaystyle \Rightarrow v=1
\displaystyle \text{Substituting }v=1\text{ in equation (iii),}
\displaystyle -2u+7=5
\displaystyle \Rightarrow u=1
\displaystyle \therefore \frac{1}{x}=1\Rightarrow x=1
\displaystyle \text{and }\frac{1}{y}=1\Rightarrow y=1.
\displaystyle \therefore x=1,\qquad y=1.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Solve the following system of equations:}
\displaystyle 152x-378y=-74,\qquad-378x+152y=-604
\displaystyle \text{Answer:}
\displaystyle 152x-378y=-74\qquad\ldots\text{(i)}
\displaystyle -378x+152y=-604\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle -226x-226y=-678
\displaystyle \Rightarrow x+y=3\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle 530x-530y=530
\displaystyle \Rightarrow x-y=1\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=4
\displaystyle \Rightarrow x=2
\displaystyle \text{Substituting }x=2\text{ in equation (iii),}
\displaystyle 2+y=3
\displaystyle \Rightarrow y=1
\displaystyle \therefore x=2,\qquad y=1.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{Solve the following system of equations:}
\displaystyle 99x+101y=499,\qquad101x+99y=501
\displaystyle \text{Answer:}
\displaystyle 99x+101y=499\qquad\ldots\text{(i)}
\displaystyle 101x+99y=501\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 200x+200y=1000
\displaystyle \Rightarrow x+y=5\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 2x-2y=2
\displaystyle \Rightarrow x-y=1\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=6
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (iii),}
\displaystyle 3+y=5
\displaystyle \Rightarrow y=2
\displaystyle \therefore x=3,\qquad y=2.
\displaystyle \\

\displaystyle \textbf{Question 21: }\text{Solve the following system of equations:}
\displaystyle 21x+47y=110,\qquad47x+21y=162
\displaystyle \text{Answer:}
\displaystyle 21x+47y=110\qquad\ldots\text{(i)}
\displaystyle 47x+21y=162\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 68x+68y=272
\displaystyle \Rightarrow x+y=4\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 26x-26y=52
\displaystyle \Rightarrow x-y=2\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=6
\displaystyle \Rightarrow x=3
\displaystyle \text{Substituting }x=3\text{ in equation (iii),}
\displaystyle 3+y=4
\displaystyle \Rightarrow y=1
\displaystyle \therefore x=3,\qquad y=1.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{Solve the following system of equations:}
\displaystyle 30x+44y=10,\qquad40x+55y=13\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle 30x+44y=10\qquad\ldots\text{(i)}
\displaystyle 40x+55y=13\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }4,
\displaystyle 120x+176y=40\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }3,
\displaystyle 120x+165y=39\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle 11y=1
\displaystyle \Rightarrow y=\frac{1}{11}
\displaystyle \text{Substituting }y=\frac{1}{11}\text{ in equation (i),}
\displaystyle 30x+44\left(\frac{1}{11}\right)=10
\displaystyle \Rightarrow 30x+4=10
\displaystyle \Rightarrow 30x=6
\displaystyle \Rightarrow x=\frac{1}{5}
\displaystyle \therefore x=\frac{1}{5},\qquad y=\frac{1}{11}.
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Solve the following system of equations:}
\displaystyle 37x+63y=137,\qquad63x+37y=163\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle 37x+63y=137\qquad\ldots\text{(i)}
\displaystyle 63x+37y=163\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 100x+100y=300
\displaystyle \Rightarrow x+y=3\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle 26x-26y=26
\displaystyle \Rightarrow x-y=1\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=4
\displaystyle \Rightarrow x=2
\displaystyle \text{Substituting }x=2\text{ in equation (iii),}
\displaystyle 2+y=3
\displaystyle \Rightarrow y=1
\displaystyle \therefore x=2,\qquad y=1.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Solve the following system of equations:}
\displaystyle 73x-37y=109,\qquad37x-73y=1\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle 73x-37y=109\qquad\ldots\text{(i)}
\displaystyle 37x-73y=1\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 110x-110y=110
\displaystyle \Rightarrow x-y=1\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (ii) from equation (i),}
\displaystyle 36x+36y=108
\displaystyle \Rightarrow x+y=3\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=4
\displaystyle \Rightarrow x=2
\displaystyle \text{Substituting }x=2\text{ in equation (iv),}
\displaystyle 2+y=3
\displaystyle \Rightarrow y=1
\displaystyle \therefore x=2,\qquad y=1.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{Solve the following system of equations:}
\displaystyle \sqrt{2}x+\sqrt{3}y=5,\qquad\sqrt{3}x-\sqrt{8}y=-\sqrt{6}\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle \sqrt{2}x+\sqrt{3}y=5\qquad\ldots\text{(i)}
\displaystyle \sqrt{3}x-\sqrt{8}y=-\sqrt{6}\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }\sqrt{3},
\displaystyle \sqrt{6}x+3y=5\sqrt{3}\qquad\ldots\text{(iii)}
\displaystyle \text{Multiplying equation (ii) by }\sqrt{2},
\displaystyle \sqrt{6}x-4y=-2\sqrt{3}\qquad\ldots\text{(iv)}
\displaystyle \text{Subtracting equation (iv) from equation (iii),}
\displaystyle 7y=7\sqrt{3}
\displaystyle \Rightarrow y=\sqrt{3}
\displaystyle \text{Substituting }y=\sqrt{3}\text{ in equation (i),}
\displaystyle \sqrt{2}x+\sqrt{3}\times\sqrt{3}=5
\displaystyle \Rightarrow \sqrt{2}x+3=5
\displaystyle \Rightarrow \sqrt{2}x=2
\displaystyle \Rightarrow x=\sqrt{2}
\displaystyle \therefore x=\sqrt{2},\qquad y=\sqrt{3}.
\displaystyle \\

\displaystyle \textbf{Question 26: }\text{Solve the following system of equations:}
\displaystyle 101x+102y=304,\qquad102x+101y=305\hfill\text{[CBSE 2025]}
\displaystyle \text{Answer:}
\displaystyle 101x+102y=304\qquad\ldots\text{(i)}
\displaystyle 102x+101y=305\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 203x+203y=609
\displaystyle \Rightarrow x+y=3\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (i) from equation (ii),}
\displaystyle x-y=1\qquad\ldots\text{(iv)}
\displaystyle \text{Adding equations (iii) and (iv),}
\displaystyle 2x=4
\displaystyle \Rightarrow x=2
\displaystyle \text{Substituting }x=2\text{ in equation (iii),}
\displaystyle 2+y=3
\displaystyle \Rightarrow y=1
\displaystyle \therefore x=2,\qquad y=1.
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{If }x+1\text{ is a factor of }2x^3+ax^2+2bx+1\text{, then find the values of }a\text{ and }b
\displaystyle \text{given that }2a-3b=4. 
\displaystyle \text{Answer:}
\displaystyle \text{Let }p(x)=2x^3+ax^2+2bx+1.
\displaystyle \text{Since }x+1\text{ is a factor, by the Factor Theorem, }p(-1)=0.
\displaystyle 2(-1)^3+a(-1)^2+2b(-1)+1=0
\displaystyle -2+a-2b+1=0
\displaystyle \Rightarrow a-2b=1\qquad\ldots\text{(i)}
\displaystyle \text{Also, }2a-3b=4\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 2a-4b=2\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (ii),}
\displaystyle b=2
\displaystyle \text{Substituting }b=2\text{ in equation (i),}
\displaystyle a-4=1
\displaystyle \Rightarrow a=5
\displaystyle \therefore a=5,\qquad b=2.
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Find the solution of the pair of equations }\frac{x}{10}+\frac{y}{5}-1=0\text{ and }\frac{x}{8}+\frac{y}{6}=15.
\displaystyle \text{Hence, find }\lambda\text{ if }y=\lambda x+5. 
\displaystyle \text{Answer:}
\displaystyle \frac{x}{10}+\frac{y}{5}-1=0
\displaystyle \Rightarrow x+2y=10\qquad\ldots\text{(i)}
\displaystyle \frac{x}{8}+\frac{y}{6}=15
\displaystyle \Rightarrow 3x+4y=360\qquad\ldots\text{(ii)}
\displaystyle \text{Multiplying equation (i) by }2,
\displaystyle 2x+4y=20\qquad\ldots\text{(iii)}
\displaystyle \text{Subtracting equation (iii) from equation (ii),}
\displaystyle x=340
\displaystyle \text{Substituting }x=340\text{ in equation (i),}
\displaystyle 340+2y=10
\displaystyle \Rightarrow 2y=-330
\displaystyle \Rightarrow y=-165
\displaystyle \therefore x=340,\qquad y=-165.
\displaystyle \text{Given }y=\lambda x+5,
\displaystyle -165=340\lambda+5
\displaystyle \Rightarrow 340\lambda=-170
\displaystyle \Rightarrow \lambda=-\frac{1}{2}
\displaystyle \therefore \lambda=-\frac{1}{2}.
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Write an equation of a line passing through the point representing}
\displaystyle \text{solution of the pair of linear equations }x+y=2\text{ and }2x-y=1.\text{ How many such lines can we find?}
\displaystyle \text{Answer:}
\displaystyle x+y=2\qquad\ldots\text{(i)}
\displaystyle 2x-y=1\qquad\ldots\text{(ii)}
\displaystyle \text{Adding equations (i) and (ii),}
\displaystyle 3x=3
\displaystyle \Rightarrow x=1
\displaystyle \text{Substituting }x=1\text{ in equation (i),}
\displaystyle 1+y=2
\displaystyle \Rightarrow y=1
\displaystyle \therefore \text{The point representing the solution is }(1,1).
\displaystyle \text{One line passing through }(1,1)\text{ is }x+y=2.
\displaystyle \text{Infinitely many lines can pass through a given point.}
\displaystyle \therefore \text{Infinitely many such lines can be found.}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Write a pair of linear equations which has the unique solution }
\displaystyle x=-1,\ y=3.  \ \text{How many such pairs can you write?} 
\displaystyle \text{Answer:}
\displaystyle \text{A pair of linear equations having the solution }x=-1,\ y=3\text{ is}
\displaystyle x+y=2\qquad\ldots\text{(i)}
\displaystyle 2x+y=1\qquad\ldots\text{(ii)}
\displaystyle \text{For these equations,}
\displaystyle \frac{a_1}{a_2}=\frac{1}{2}\ne\frac{1}{1}=\frac{b_1}{b_2}
\displaystyle \therefore \text{The pair has a unique solution.}
\displaystyle \text{Also, }(-1)+3=2\text{ and }2(-1)+3=1.
\displaystyle \therefore x=-1,\ y=3\text{ is the unique solution of the pair.}
\displaystyle \text{Infinitely many such pairs of linear equations can be written.}
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.