\displaystyle \textbf{Question 1: }\text{The pair of equations }y=0\text{ and }y=-7\text{ has}
\displaystyle \underline{\hspace{1.5cm}}\text{ solution.}
\displaystyle \text{Answer:}
\displaystyle \text{The lines }y=0\text{ and }y=-7\text{ are parallel.}
\displaystyle \therefore \text{The pair of equations has }\textbf{no}\text{ solution.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The pair of equations }x=a\text{ and }y=b\text{ has}
\displaystyle \underline{\hspace{1.5cm}}\text{ solution.}
\displaystyle \text{Answer:}
\displaystyle \text{The lines }x=a\text{ and }y=b\text{ intersect at }(a,b).
\displaystyle \therefore \text{The pair of equations has a }\textbf{unique}\text{ solution.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{If the pair of equations }3x+2ky=2\text{ and}
\displaystyle 2x+5y+1=0\text{ has no solution, then }k=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \frac{3}{2}=\frac{2k}{5}\ne\frac{-2}{1}
\displaystyle \Rightarrow 4k=15\Rightarrow k=\frac{15}{4}.
\displaystyle \therefore \text{The required value is }\frac{15}{4}.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{If a pair of linear equations is consistent, then}
\displaystyle \text{the lines representing them are either }\underline{\hspace{1.2cm}}\text{ or }\underline{\hspace{1.2cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{A consistent pair has either one solution or infinitely many solutions.}
\displaystyle \therefore \text{The lines are either }\textbf{intersecting}\text{ or }\textbf{coincident}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{There is (are) }\underline{\hspace{1.2cm}}\text{ value(s) of }c\text{ for which}
\displaystyle \text{the pair of equations }cx-y=2\text{ and }6x-2y=3\text{ have infinitely}
\displaystyle \text{many solutions.}
\displaystyle \text{Answer:}
\displaystyle \text{For infinitely many solutions, }\frac{c}{6}=\frac{-1}{-2}=\frac{-2}{-3}.
\displaystyle \text{But }\frac{1}{2}\ne\frac{2}{3}.
\displaystyle \therefore \text{There is }\textbf{no value}\text{ of }c.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If a pair of linear equations is consistent with a}
\displaystyle \text{unique solution, then the lines representing them are }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{A consistent pair with a unique solution represents intersecting lines.}
\displaystyle \therefore \text{The required answer is }\textbf{intersecting}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The pair of equations }\lambda x+3y=7,\ 2x+6y=14
\displaystyle \text{will have infinitely many solutions for }\lambda=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{\lambda}{2}=\frac{3}{6}=\frac{7}{14}
\displaystyle \Rightarrow \frac{\lambda}{2}=\frac{1}{2}\Rightarrow\lambda=1.
\displaystyle \therefore \text{The required value is }1.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{If the pair of equations }2x+3y-5=0\text{ and}
\displaystyle px-6y-8=0\text{ has a unique solution for all real values of }p\text{ except}
\displaystyle \underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For a unique solution, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.
\displaystyle \therefore \frac{2}{p}\ne\frac{3}{-6}
\displaystyle \Rightarrow \frac{2}{p}\ne-\frac{1}{2}\Rightarrow p\ne-4.
\displaystyle \therefore \text{The required value is }-4.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{If the lines represented by the equations}
\displaystyle 3x-y-5=0\text{ and }6x-2y-p=0\text{ are parallel, then }p\text{ is equal to}
\displaystyle \underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For parallel lines, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \frac{3}{6}=\frac{-1}{-2}\ne\frac{-5}{-p}
\displaystyle \Rightarrow \frac{1}{2}\ne\frac{5}{p}\Rightarrow p\ne10.
\displaystyle \therefore p\ne10.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{If }x=a,\ y=b\text{ is the solution of the pair of}
\displaystyle \text{equations }x-y=2\text{ and }x+y=4,\text{ then }a=\underline{\hspace{1.2cm}}
\displaystyle \text{and }b=\underline{\hspace{1.2cm}}.
\displaystyle \text{Answer:}
\displaystyle x-y=2,\qquad x+y=4
\displaystyle \text{Adding the equations, }2x=6\Rightarrow x=3.
\displaystyle \therefore y=4-3=1.
\displaystyle \therefore a=3,\quad b=1.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{If one equation of a pair of dependent linear}
\displaystyle \text{equations is }5x-7y+2=0,\text{ then the second equation is given by}
\displaystyle \underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{Dependent equations represent the same line.}
\displaystyle \text{Hence, the second equation is any non-zero multiple of the first.}
\displaystyle \text{For example, multiplying by }2,
\displaystyle \therefore 10x-14y+4=0.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{If the pair of equations }ax+2y=7\text{ and}
\displaystyle 3x+by=16\text{ represent parallel lines, then }ab=\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{For parallel lines, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \frac{a}{3}=\frac{2}{b}
\displaystyle \Rightarrow ab=6.
\displaystyle \therefore ab=6.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The line }4x+3y-12=0\text{ cuts the coordinate}
\displaystyle \text{axes at }A\text{ and }B.\text{ The area of }\triangle OAB\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle y=0\Rightarrow x=3,\qquad x=0\Rightarrow y=4.
\displaystyle \text{Area of }\triangle OAB=\frac{1}{2}\times3\times4=6\text{ sq. units.}
\displaystyle \therefore \text{The required answer is }6\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{If }2^{x+y}=2^{x-y}=\sqrt{8},\text{ then }x=\underline{\hspace{1.2cm}}
\displaystyle \text{and }y=\underline{\hspace{1.2cm}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{8}=2^{3/2}
\displaystyle \therefore x+y=\frac{3}{2},\qquad x-y=\frac{3}{2}.
\displaystyle \text{Adding, }2x=3\Rightarrow x=\frac{3}{2},\qquad y=0.
\displaystyle \therefore x=\frac{3}{2},\quad y=0.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The system of equations }ax+3y=1,\ -12x+ay=2
\displaystyle \text{has }\underline{\hspace{1.5cm}}\text{ for all real values of }a.
\displaystyle \text{Answer:}
\displaystyle a_1b_2-a_2b_1=a^2-(-12)(3)=a^2+36>0.
\displaystyle \therefore \text{The system has a }\textbf{unique solution}\text{ for all real values of }a.
\displaystyle \\

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