Class 10: Linear Equations – Very Short Answer Type Questions

$\displaystyle \textbf{Question 1: }\text{Write the value of }k\text{ for which the system of}$
$\displaystyle \text{equations }x+y-4=0\text{ and }2x+ky-3=0\text{ has no solution.}$
$\displaystyle \hfill\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.$
$\displaystyle \therefore \frac{1}{2}=\frac{1}{k}\ne\frac{-4}{-3}$
$\displaystyle \Rightarrow k=2$
$\displaystyle \therefore k=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Write the value of }k\text{ for which the system of}$
$\displaystyle \text{equations }2x-y=5\text{ and }6x+ky=15\text{ has infinitely many}$
$\displaystyle \text{solutions.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$
$\displaystyle \therefore \frac{2}{6}=\frac{-1}{k}=\frac{-5}{-15}$
$\displaystyle \Rightarrow \frac{1}{3}=\frac{-1}{k}$
$\displaystyle \Rightarrow k=-3$
$\displaystyle \therefore k=-3.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Write the value of }k\text{ for which the system of}$
$\displaystyle \text{equations }3x-2y=0\text{ and }kx+5y=0\text{ has infinitely many solutions.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}.$
$\displaystyle \therefore \frac{3}{k}=\frac{-2}{5}$
$\displaystyle \Rightarrow 15=-2k$
$\displaystyle \Rightarrow k=-\frac{15}{2}$
$\displaystyle \therefore k=-\frac{15}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Write the values of }k\text{ for which the system of}$
$\displaystyle \text{equations }x+ky=0\text{ and }2x-y=0\text{ has a unique solution.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a unique solution, }\frac{a_1}{a_2}\ne\frac{b_1}{b_2}.$
$\displaystyle \therefore \frac{1}{2}\ne\frac{k}{-1}$
$\displaystyle \Rightarrow k\ne-\frac{1}{2}$
$\displaystyle \therefore k\ne-\frac{1}{2}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Write the set of values of }a\text{ and }b\text{ for which the}$
$\displaystyle \text{following system of equations has infinitely many solutions.}$
$\displaystyle 2x+3y=7$
$\displaystyle 2ax+(a+b)y=28$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$
$\displaystyle \therefore \frac{2}{2a}=\frac{3}{a+b}=\frac{7}{28}$
$\displaystyle \frac{1}{a}=\frac{1}{4}\Rightarrow a=4$
$\displaystyle \frac{3}{a+b}=\frac{1}{4}\Rightarrow a+b=12$
$\displaystyle \Rightarrow b=8$
$\displaystyle \therefore a=4,\quad b=8.$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{For what value of }k\text{, the following pair of linear}$
$\displaystyle \text{equations has infinitely many solutions?}$
$\displaystyle 10x+5y-(k-5)=0$
$\displaystyle 20x+10y-k=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For infinitely many solutions, }\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}.$
$\displaystyle \therefore \frac{10}{20}=\frac{5}{10}=\frac{5-k}{-k}$
$\displaystyle \Rightarrow \frac{1}{2}=\frac{5-k}{-k}$
$\displaystyle \Rightarrow -k=10-2k$
$\displaystyle \Rightarrow k=10$
$\displaystyle \therefore k=10.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Write the number of solutions of the following pair}$
$\displaystyle \text{of linear equations:}\hfill\text{[CBSE 2009]}$
$\displaystyle x+2y-8=0$
$\displaystyle 2x+4y=16$
$\displaystyle \text{Answer:}$
$\displaystyle 2x+4y=16\Rightarrow x+2y-8=0$
$\displaystyle \text{Thus, both equations represent the same line.}$
$\displaystyle \therefore \text{The pair of equations has infinitely many solutions.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Write the number of solutions of the following pair}$
$\displaystyle \text{of linear equations:}$
$\displaystyle x+3y-4=0$
$\displaystyle 2x+6y=7$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For no solution, }\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}.$
$\displaystyle \frac{1}{2}=\frac{3}{6}\ne\frac{-4}{-7}$
$\displaystyle \therefore \text{The pair of equations has no solution.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }2x+y=13\text{ and }4x-y=17,\text{ find the value}$
$\displaystyle \text{of }(x-y).\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle 2x+y=13\qquad\ldots\text{(i)}$
$\displaystyle 4x-y=17\qquad\ldots\text{(ii)}$
$\displaystyle \text{Adding equations (i) and (ii), }6x=30\Rightarrow x=5.$
$\displaystyle \text{From equation (i), }10+y=13\Rightarrow y=3.$
$\displaystyle \therefore x-y=5-3=2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Sum of two numbers is }105\text{ and their difference}$
$\displaystyle \text{is }45.\text{ Find the numbers.}\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two numbers be }x\text{ and }y,\text{ where }x>y.$
$\displaystyle x+y=105,\qquad x-y=45$
$\displaystyle \text{Adding, }2x=150\Rightarrow x=75.$
$\displaystyle \therefore y=105-75=30.$
$\displaystyle \therefore \text{The numbers are }75\text{ and }30.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Solve for }x\text{ and }y:$
$\displaystyle 23x+24y=23$
$\displaystyle 24x+23y=24\hfill\text{[CBSE 2024]}$
$\displaystyle \text{Answer:}$
$\displaystyle 23x+24y=23\qquad\ldots\text{(i)}$
$\displaystyle 24x+23y=24\qquad\ldots\text{(ii)}$
$\displaystyle \text{Subtracting equation (i) from equation (ii), }x-y=1.$
$\displaystyle \text{Adding equations (i) and (ii), }47x+47y=47.$
$\displaystyle \Rightarrow x+y=1$
$\displaystyle \therefore x=1,\quad y=0.$
$\displaystyle \\$