\displaystyle \textbf{Question 1: }\text{Determine whether the following system has a unique solution, no solution or}
\displaystyle \text{infinitely many solutions. In case there is a unique solution, find it:}
\displaystyle x-3y=3,\qquad3x-9y=2
\displaystyle \text{Answer:}
\displaystyle x-3y-3=0\qquad\ldots\text{(i)}
\displaystyle 3x-9y-2=0\qquad\ldots\text{(ii)}
\displaystyle a_1=1,\quad b_1=-3,\quad c_1=-3
\displaystyle a_2=3,\quad b_2=-9,\quad c_2=-2
\displaystyle \frac{a_1}{a_2}=\frac{1}{3}
\displaystyle \frac{b_1}{b_2}=\frac{-3}{-9}=\frac{1}{3}
\displaystyle \frac{c_1}{c_2}=\frac{-3}{-2}=\frac{3}{2}
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}
\displaystyle \therefore \text{The given system is inconsistent and has no solution.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Determine whether the following system has a unique solution, no solution or}
\displaystyle \text{infinitely many solutions. In case there is a unique solution, find it:}
\displaystyle 2x+y=5,\qquad4x+2y=10
\displaystyle \text{Answer:}
\displaystyle 2x+y-5=0\qquad\ldots\text{(i)}
\displaystyle 4x+2y-10=0\qquad\ldots\text{(ii)}
\displaystyle a_1=2,\quad b_1=1,\quad c_1=-5
\displaystyle a_2=4,\quad b_2=2,\quad c_2=-10
\displaystyle \frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}
\displaystyle \frac{b_1}{b_2}=\frac{1}{2}
\displaystyle \frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}
\displaystyle \therefore \text{The given system is consistent and has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the value of }k\text{ for which the following system of equations has a unique solution:}
\displaystyle kx+2y=5,\qquad3x+y=1
\displaystyle \text{Answer:}
\displaystyle kx+2y-5=0\qquad\ldots\text{(i)}
\displaystyle 3x+y-1=0\qquad\ldots\text{(ii)}
\displaystyle a_1=k,\quad b_1=2,\quad a_2=3,\quad b_2=1
\displaystyle \text{For a unique solution, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.
\displaystyle \therefore \frac{k}{3}\ne\frac{2}{1}
\displaystyle \Rightarrow k\ne6
\displaystyle \therefore \text{The given system has a unique solution for all real values of }k\text{ other than }6.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the value of }k\text{ for which the following system of equations has a unique solution:}
\displaystyle 4x+ky+8=0,\qquad2x+2y+2=0 
\displaystyle \text{Answer:}
\displaystyle 4x+ky+8=0\qquad\ldots\text{(i)}
\displaystyle 2x+2y+2=0\qquad\ldots\text{(ii)}
\displaystyle a_1=4,\quad b_1=k,\quad a_2=2,\quad b_2=2
\displaystyle \text{For a unique solution, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.
\displaystyle \therefore \frac{4}{2}\ne\frac{k}{2}
\displaystyle \Rightarrow 2\ne\frac{k}{2}
\displaystyle \Rightarrow k\ne4
\displaystyle \therefore \text{The given system has a unique solution for all real values of }k\text{ other than }4.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the value of }k\text{ for which the following system of equations has}
\displaystyle \text{infinitely many solutions: }2x+3y=2, \ (k+2)x+(2k+1)y=2(k-1)\hfill\text{[CBSE 2000, 2003]}
\displaystyle \text{Answer:}
\displaystyle 2x+3y-2=0\qquad\ldots\text{(i)}
\displaystyle (k+2)x+(2k+1)y-2(k-1)=0\qquad\ldots\text{(ii)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{k+2}=\frac{3}{2k+1}=\frac{-2}{-2(k-1)}
\displaystyle \frac{2}{k+2}=\frac{3}{2k+1}
\displaystyle \Rightarrow 4k+2=3k+6
\displaystyle \Rightarrow k=4
\displaystyle \text{For }k=4,\quad\frac{2}{6}=\frac{3}{9}=\frac{-2}{-6}=\frac{1}{3}.
\displaystyle \therefore \text{The given system has infinitely many solutions for }k=4.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the value of }k\text{ for which the following system of equations has}
\displaystyle \text{infinitely many solutions: }x+(k+1)y=4,\qquad(k+1)x+9y=5k+2\hfill\text{[CBSE 2000C]}
\displaystyle \text{Answer:}
\displaystyle x+(k+1)y-4=0\qquad\ldots\text{(i)}
\displaystyle (k+1)x+9y-(5k+2)=0\qquad\ldots\text{(ii)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{1}{k+1}=\frac{k+1}{9}=\frac{-4}{-(5k+2)}
\displaystyle \frac{1}{k+1}=\frac{k+1}{9}
\displaystyle \Rightarrow (k+1)^2=9
\displaystyle \Rightarrow k+1=\pm3
\displaystyle \Rightarrow k=2\text{ or }k=-4
\displaystyle \text{For }k=2,\quad\frac{1}{3}=\frac{3}{9}=\frac{-4}{-12}=\frac{1}{3}.
\displaystyle \therefore k=2\text{ satisfies all three ratios.}
\displaystyle \text{For }k=-4,\quad\frac{1}{-3}=\frac{-3}{9}\ne\frac{-4}{18}.
\displaystyle \therefore k=-4\text{ does not satisfy all three ratios.}
\displaystyle \therefore \text{The given system has infinitely many solutions for }k=2.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find the value of }k\text{ for which the following system of equations has}
\displaystyle \text{infinitely many solutions: }2x+(k-2)y=k, \ 6x+(2k-1)y=2k+5\hfill\text{[CBSE 2000C]}
\displaystyle \text{Answer:}
\displaystyle 2x+(k-2)y-k=0\qquad\ldots\text{(i)}
\displaystyle 6x+(2k-1)y-(2k+5)=0\qquad\ldots\text{(ii)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{6}=\frac{k-2}{2k-1}=\frac{-k}{-(2k+5)}
\displaystyle \frac{1}{3}=\frac{k-2}{2k-1}
\displaystyle \Rightarrow 2k-1=3k-6
\displaystyle \Rightarrow k=5
\displaystyle \text{For }k=5,\quad\frac{2}{6}=\frac{3}{9}=\frac{-5}{-15}=\frac{1}{3}.
\displaystyle \therefore \text{The given system has infinitely many solutions for }k=5.
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Find the value of }k\text{ for which the following system of equations has}
\displaystyle \text{infinitely many solutions: }2x+3y=7, \ (k+1)x+(2k-1)y=4k+1\hfill\text{[CBSE 2001]}
\displaystyle \text{Answer:}
\displaystyle 2x+3y-7=0\qquad\ldots\text{(i)}
\displaystyle (k+1)x+(2k-1)y-(4k+1)=0\qquad\ldots\text{(ii)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{k+1}=\frac{3}{2k-1}=\frac{-7}{-(4k+1)}
\displaystyle \frac{2}{k+1}=\frac{3}{2k-1}
\displaystyle \Rightarrow 2(2k-1)=3(k+1)
\displaystyle \Rightarrow 4k-2=3k+3
\displaystyle \Rightarrow k=5
\displaystyle \text{For }k=5,\quad\frac{2}{6}=\frac{3}{9}=\frac{-7}{-21}=\frac{1}{3}.
\displaystyle \therefore \text{The given system has infinitely many solutions for }k=5.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Find the value of }k\text{ for which the following system of equations has no solution:}
\displaystyle 2x+ky=11,\qquad5x-7y=5
\displaystyle \text{Answer:}
\displaystyle 2x+ky-11=0\qquad\ldots\text{(i)}
\displaystyle 5x-7y-5=0\qquad\ldots\text{(ii)}
\displaystyle a_1=2,\quad b_1=k,\quad c_1=-11
\displaystyle a_2=5,\quad b_2=-7,\quad c_2=-5
\displaystyle \text{For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{5}=\frac{k}{-7}
\displaystyle \Rightarrow 14=-5k
\displaystyle \Rightarrow k=-\frac{14}{5}
\displaystyle \text{Also, }\frac{c_1}{c_2}=\frac{-11}{-5}=\frac{11}{5}\ne\frac{2}{5}.
\displaystyle \therefore \text{The given system has no solution for }k=-\frac{14}{5}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the value of }k\text{ for which the following system of equations has no solution:}
\displaystyle kx+3y=k-3,\qquad12x+ky=6
\displaystyle \text{Answer:}
\displaystyle kx+3y-k+3=0\qquad\ldots\text{(i)}
\displaystyle 12x+ky-6=0\qquad\ldots\text{(ii)}
\displaystyle a_1=k,\quad b_1=3,\quad c_1=3-k
\displaystyle a_2=12,\quad b_2=k,\quad c_2=-6
\displaystyle \text{For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \frac{k}{12}=\frac{3}{k}
\displaystyle \Rightarrow k^2=36
\displaystyle \Rightarrow k=6\text{ or }k=-6
\displaystyle \text{For }k=6,\quad\frac{a_1}{a_2}=\frac{6}{12}=\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{and }\frac{c_1}{c_2}=\frac{3-6}{-6}=\frac{1}{2}.
\displaystyle \therefore k=6\text{ gives infinitely many solutions, not no solution.}
\displaystyle \text{For }k=-6,\quad\frac{a_1}{a_2}=\frac{-6}{12}=-\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{3}{-6}=-\frac{1}{2}
\displaystyle \text{and }\frac{c_1}{c_2}=\frac{3-(-6)}{-6}=-\frac{3}{2}.
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \text{The given system has no solution for }k=-6.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Find the value of }k\text{ for which the following system of equations has no solution:}
\displaystyle kx+3y=k-3,\qquad12x+ky=k\hfill\text{[CBSE 2003, 2009]}
\displaystyle \text{Answer:}
\displaystyle kx+3y-k+3=0\qquad\ldots\text{(i)}
\displaystyle 12x+ky-k=0\qquad\ldots\text{(ii)}
\displaystyle a_1=k,\quad b_1=3,\quad c_1=3-k
\displaystyle a_2=12,\quad b_2=k,\quad c_2=-k
\displaystyle \text{For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \frac{k}{12}=\frac{3}{k}
\displaystyle \Rightarrow k^2=36
\displaystyle \Rightarrow k=6\text{ or }k=-6
\displaystyle \text{For }k=6,\quad\frac{a_1}{a_2}=\frac{6}{12}=\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}
\displaystyle \text{and }\frac{c_1}{c_2}=\frac{3-6}{-6}=\frac{1}{2}.
\displaystyle \therefore k=6\text{ gives infinitely many solutions, not no solution.}
\displaystyle \text{For }k=-6,\quad\frac{a_1}{a_2}=\frac{-6}{12}=-\frac{1}{2},\quad\frac{b_1}{b_2}=\frac{3}{-6}=-\frac{1}{2}
\displaystyle \text{and }\frac{c_1}{c_2}=\frac{3-(-6)}{6}=\frac{3}{2}.
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \text{The given system has no solution for }k=-6.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Find }c\text{ if the following system of equations has infinitely many solutions:}
\displaystyle cx+3y+3-c=0,\qquad12x+cy-c=0\hfill\text{[CBSE 2019]}
\displaystyle \text{Answer:}
\displaystyle a_1=c,\quad b_1=3,\quad c_1=3-c
\displaystyle a_2=12,\quad b_2=c,\quad c_2=-c
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{c}{12}=\frac{3}{c}=\frac{3-c}{-c}
\displaystyle \frac{c}{12}=\frac{3}{c}
\displaystyle \Rightarrow c^2=36
\displaystyle \Rightarrow c=6\text{ or }c=-6
\displaystyle \text{For }c=6,\quad\frac{6}{12}=\frac{3}{6}=\frac{3-6}{-6}=\frac{1}{2}.
\displaystyle \therefore c=6\text{ satisfies all three ratios.}
\displaystyle \text{For }c=-6,\quad\frac{-6}{12}=\frac{3}{-6}=-\frac{1}{2}\ne\frac{3-(-6)}{6}=\frac{3}{2}.
\displaystyle \therefore c=-6\text{ does not satisfy all three ratios.}
\displaystyle \therefore \text{The given system has infinitely many solutions for }c=6.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find the values of }k\text{ for which the system }2x+ky=1,\ 
\displaystyle 3x-5y=7\text{ will have} \ \text{(i) a unique solution, and (ii) no solution. Is there a value of }
\displaystyle k\text{ for which the system has} \ \text{infinitely many solutions?}
\displaystyle \text{Answer:}
\displaystyle 2x+ky-1=0\qquad\ldots\text{(i)}
\displaystyle 3x-5y-7=0\qquad\ldots\text{(ii)}
\displaystyle a_1=2,\quad b_1=k,\quad c_1=-1,\quad a_2=3,\quad b_2=-5,\quad c_2=-7

\displaystyle \text{(i) For a unique solution, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.
\displaystyle \therefore \frac{2}{3}\ne\frac{k}{-5}
\displaystyle \Rightarrow 10\ne-3k
\displaystyle \Rightarrow k\ne-\frac{10}{3}
\displaystyle \therefore \text{The system has a unique solution for all real values of }k\ne-\frac{10}{3}.

\displaystyle \text{(ii) For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{3}=\frac{k}{-5}
\displaystyle \Rightarrow 10=-3k
\displaystyle \Rightarrow k=-\frac{10}{3}
\displaystyle \text{Also, }\frac{c_1}{c_2}=\frac{-1}{-7}=\frac{1}{7}\ne\frac{2}{3}.
\displaystyle \therefore \text{The system has no solution for }k=-\frac{10}{3}.

\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \text{But }\frac{a_1}{a_2}=\frac{2}{3}\ne\frac{1}{7}=\frac{c_1}{c_2}.
\displaystyle \therefore \text{There is no value of }k\text{ for which the system has infinitely many solutions.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{For what value of }k\text{, will the following system of equations represent the}
\displaystyle \text{coincident lines?}\qquad x+2y+7=0,\qquad2x+ky+14=0
\displaystyle \text{Answer:}
\displaystyle x+2y+7=0\qquad\ldots\text{(i)}
\displaystyle 2x+ky+14=0\qquad\ldots\text{(ii)}
\displaystyle a_1=1,\quad b_1=2,\quad c_1=7,\quad a_2=2,\quad b_2=k,\quad c_2=14
\displaystyle \text{For coincident lines, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{1}{2}=\frac{2}{k}=\frac{7}{14}
\displaystyle \frac{1}{2}=\frac{2}{k}
\displaystyle \Rightarrow k=4
\displaystyle \therefore \text{The given equations represent coincident lines for }k=4.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{Determine the values of }a\text{ and }b\text{ so that the following system of linear equations}
\displaystyle \text{has infinitely many solutions:}
\displaystyle (2a-1)x+3y-5=0,\qquad3x+(b-1)y-2=0\hfill\text{[CBSE 2001C]}
\displaystyle \text{Answer:}
\displaystyle a_1=2a-1,\quad b_1=3,\quad c_1=-5
\displaystyle a_2=3,\quad b_2=b-1,\quad c_2=-2
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2a-1}{3}=\frac{3}{b-1}=\frac{-5}{-2}=\frac{5}{2}
\displaystyle \frac{2a-1}{3}=\frac{5}{2}
\displaystyle \Rightarrow 4a-2=15
\displaystyle \Rightarrow 4a=17
\displaystyle \Rightarrow a=\frac{17}{4}
\displaystyle \frac{3}{b-1}=\frac{5}{2}
\displaystyle \Rightarrow 6=5(b-1)
\displaystyle \Rightarrow 6=5b-5
\displaystyle \Rightarrow 5b=11
\displaystyle \Rightarrow b=\frac{11}{5}
\displaystyle \therefore a=\frac{17}{4},\qquad b=\frac{11}{5}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{For which value(s) of }\lambda\text{, do the pair of linear equations }
\displaystyle \lambda x+y=\lambda^2\text{ and}  \ x+\lambda y=1 \\ \text{have (i) no solution? (ii) infinitely many solutions? (iii) a unique solution?}
\displaystyle   
\displaystyle \text{Answer:}
\displaystyle \lambda x+y-\lambda^2=0\qquad\ldots\text{(i)}
\displaystyle x+\lambda y-1=0\qquad\ldots\text{(ii)}
\displaystyle a_1=\lambda,\quad b_1=1,\quad c_1=-\lambda^2
\displaystyle a_2=1,\quad b_2=\lambda,\quad c_2=-1

\displaystyle \text{(i) For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \lambda=\frac{1}{\lambda}
\displaystyle \Rightarrow \lambda^2=1
\displaystyle \Rightarrow \lambda=1\text{ or }\lambda=-1
\displaystyle \text{For }\lambda=1,\quad\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=1.
\displaystyle \therefore \lambda=1\text{ does not give no solution.}
\displaystyle \text{For }\lambda=-1,\quad\frac{a_1}{a_2}=-1,\quad\frac{b_1}{b_2}=-1,\quad\frac{c_1}{c_2}=1.
\displaystyle \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.
\displaystyle \therefore \text{The pair has no solution for }\lambda=-1.

\displaystyle \text{(ii) For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \text{For }\lambda=1,\quad\frac{1}{1}=\frac{1}{1}=\frac{-1}{-1}=1.
\displaystyle \therefore \text{The pair has infinitely many solutions for }\lambda=1.

\displaystyle \text{(iii) For a unique solution, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.
\displaystyle \therefore \lambda\ne\frac{1}{\lambda}
\displaystyle \Rightarrow \lambda^2\ne1
\displaystyle \Rightarrow \lambda\ne\pm1
\displaystyle \therefore \text{The pair has a unique solution for }\lambda\ne\pm1.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the values of }a\text{ and }b\text{ for which the following system }
\displaystyle \text{of equations has infinitely many solutions:}
\displaystyle \text{(i) }(2a-1)x-3y=5,\qquad3x+(b-2)y=3\hfill\text{[CBSE 2002C]}
\displaystyle \text{(ii) }2x-3y=7,\qquad(a+b)x-(a+b-3)y=4a+b\hfill\text{[CBSE 2002C]}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }(2a-1)x-3y-5=0\qquad\ldots\text{(1)}
\displaystyle 3x+(b-2)y-3=0\qquad\ldots\text{(2)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2a-1}{3}=\frac{-3}{b-2}=\frac{-5}{-3}=\frac{5}{3}
\displaystyle \frac{2a-1}{3}=\frac{5}{3}
\displaystyle \Rightarrow 2a-1=5
\displaystyle \Rightarrow a=3
\displaystyle \frac{-3}{b-2}=\frac{5}{3}
\displaystyle \Rightarrow -9=5b-10
\displaystyle \Rightarrow b=\frac{1}{5}
\displaystyle \therefore a=3,\qquad b=\frac{1}{5}.

\displaystyle \text{(ii) }2x-3y-7=0\qquad\ldots\text{(1)}
\displaystyle (a+b)x-(a+b-3)y-(4a+b)=0\qquad\ldots\text{(2)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{a+b}=\frac{-3}{-(a+b-3)}=\frac{-7}{-(4a+b)}
\displaystyle \frac{2}{a+b}=\frac{3}{a+b-3}
\displaystyle \Rightarrow 2(a+b-3)=3(a+b)
\displaystyle \Rightarrow a+b=-6\qquad\ldots\text{(3)}
\displaystyle \frac{2}{a+b}=\frac{7}{4a+b}
\displaystyle \Rightarrow 2(4a+b)=7(a+b)
\displaystyle \Rightarrow a-5b=0
\displaystyle \Rightarrow a=5b\qquad\ldots\text{(4)}
\displaystyle \text{Substituting }a=5b\text{ in (3),}
\displaystyle 5b+b=-6
\displaystyle \Rightarrow b=-1
\displaystyle \therefore a=-5
\displaystyle \therefore a=-5,\qquad b=-1.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the values of }a\text{ and }b\text{ for which the following system }
\displaystyle \text{of equations has infinitely many solutions:}
\displaystyle \text{(iii) }2x+3y=7,\qquad(a-b)x+(a+b)y=3a+b-2 
\displaystyle \text{(iv) }2x+3y-7=0,\qquad(a-1)x+(a+1)y=3a-1\hfill\text{[CBSE 2010]}
\displaystyle \text{Answer:}
\displaystyle \text{(iii) }2x+3y-7=0\qquad\ldots\text{(1)}
\displaystyle (a-b)x+(a+b)y-(3a+b-2)=0\qquad\ldots\text{(2)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{a-b}=\frac{3}{a+b}=\frac{-7}{-(3a+b-2)}
\displaystyle \frac{2}{a-b}=\frac{3}{a+b}
\displaystyle \Rightarrow 2(a+b)=3(a-b)
\displaystyle \Rightarrow a=5b\qquad\ldots\text{(3)}
\displaystyle \frac{2}{a-b}=\frac{7}{3a+b-2}
\displaystyle \Rightarrow 2(3a+b-2)=7(a-b)
\displaystyle \Rightarrow a-9b+4=0\qquad\ldots\text{(4)}
\displaystyle \text{Substituting }a=5b\text{ in (4),}
\displaystyle 5b-9b+4=0
\displaystyle \Rightarrow b=1
\displaystyle \therefore a=5
\displaystyle \therefore a=5,\qquad b=1.

\displaystyle \text{(iv) }2x+3y-7=0\qquad\ldots\text{(1)}
\displaystyle (a-1)x+(a+1)y-(3a-1)=0\qquad\ldots\text{(2)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{a-1}=\frac{3}{a+1}=\frac{-7}{-(3a-1)}
\displaystyle \frac{2}{a-1}=\frac{3}{a+1}
\displaystyle \Rightarrow 2(a+1)=3(a-1)
\displaystyle \Rightarrow 2a+2=3a-3
\displaystyle \Rightarrow a=5
\displaystyle \text{For }a=5,\quad\frac{2}{4}=\frac{3}{6}=\frac{-7}{-14}=\frac{1}{2}.
\displaystyle \therefore \text{The given system has infinitely many solutions for }a=5.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find the values of }a\text{ and }b\text{ for which the following system }
\displaystyle \text{of equations has infinitely many solutions:}
\displaystyle \text{(v) }x+2y=1,\qquad(a-b)x+(a+b)y=a+b-2
\displaystyle \text{(vi) }2x+3y=7,\qquad2ax+ay=28-by
\displaystyle \text{Answer:}
\displaystyle \text{(v) }x+2y-1=0\qquad\ldots\text{(1)}
\displaystyle (a-b)x+(a+b)y-(a+b-2)=0\qquad\ldots\text{(2)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{1}{a-b}=\frac{2}{a+b}=\frac{-1}{-(a+b-2)}
\displaystyle \frac{1}{a-b}=\frac{2}{a+b}
\displaystyle \Rightarrow a+b=2(a-b)
\displaystyle \Rightarrow a=3b\qquad\ldots\text{(3)}
\displaystyle \frac{2}{a+b}=\frac{1}{a+b-2}
\displaystyle \Rightarrow 2(a+b-2)=a+b
\displaystyle \Rightarrow a+b=4\qquad\ldots\text{(4)}
\displaystyle \text{Substituting }a=3b\text{ in (4),}
\displaystyle 3b+b=4
\displaystyle \Rightarrow b=1
\displaystyle \therefore a=3
\displaystyle \therefore a=3,\qquad b=1.

\displaystyle \text{(vi) }2x+3y=7,\qquad2ax+ay=28-by
\displaystyle \Rightarrow 2x+3y-7=0\qquad\ldots\text{(1)}
\displaystyle \Rightarrow 2ax+(a+b)y-28=0\qquad\ldots\text{(2)}
\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.
\displaystyle \therefore \frac{2}{2a}=\frac{3}{a+b}=\frac{-7}{-28}=\frac{1}{4}
\displaystyle \frac{2}{2a}=\frac{1}{4}
\displaystyle \Rightarrow \frac{1}{a}=\frac{1}{4}
\displaystyle \Rightarrow a=4
\displaystyle \frac{3}{a+b}=\frac{1}{4}
\displaystyle \Rightarrow a+b=12
\displaystyle \Rightarrow 4+b=12
\displaystyle \Rightarrow b=8
\displaystyle \therefore a=4,\qquad b=8.
\displaystyle \\

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

Subscribe to get the latest posts sent to your email.