NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables

Ans : Let the cost of a notebook and a pen be x and y respectively. 
Cost of notebook = 2 x Cost of pen 
\( \begin{array}{l}{x=2 y} \\ {x-2 y=0}\end{array} \) 

Ans : (i) \( 2 x+3 y=9.3 \overline{5} \) 
\( 2 x+3 y-9.35=0 \) 
\( \begin{array}{l}{\text { Comparing this equation with } a x+b y+c=0} \\ {a=2, b=3, c=-9.3 \overline{5}}\end{array} \) 
(ii) \( x-\frac{y}{5}-10=0 \) 
\( \begin{array}{l}{\text { Comparing this equation with } a x+b y+c=0} \\ {a=1, b=-\frac{1}{5}, c=-10}\end{array} \) 
(iii) \( -2 x+3 y=6 \) 
\( \begin{array}{l}{-2 x+3 y-6=0} \\ {\text { Comparing this equation with } a x+b y+c=0} \\ {a=-2, b=3, c=-6}\end{array} \) 
(iv) \( x=3 y \) 
\( \begin{array}{l}{1 x-3 y+0=0} \\ {\text { Comparing this equation with } a x+b y+c=0} \\ {a=1, b=-3, c=0}\end{array} \) 
(v) \( 2 x=-5 y \) 
\( \begin{array}{l}{2 x+5 y+0=0} \\ {\text { Comparing this equation with } a x+b y+c=0 \text { , }} \\ {a=2, b=5, c=0}\end{array} \) 
(vi) \( 3 x+2=0 \) 
\( \begin{array}{l}{3 x+0 . y+2=0} \\ {\text { Comparing this equation with } a x+b y+c=0} \\ {a=3, b=0, c=2}\end{array} \) 
(vii) \( y-2=0 \) 
\( \begin{array}{l}{0 . x+1 . y-2=0} \\ {\text { Comparing this equation with } a x+b y+c=0} \\ {a=0, b=1, c=-2}\end{array} \) 
(viii) \( 5=2 x \) 
\( \begin{array}{l}{-2 x+0 . y+5=0} \\ {\text { Comparing this equation with } a x+b y+c=0} \\ {a=-2, b=0, c=5}\end{array} \) 

Ans : y = 3x + 5 is a linear equation in two variables and it has infinite possible solutions. 
As for every value of x, there will be a value of y satisfying the above equation and 
vice-versa. 
Hence, the correct answer is (iii). 

Ans : (i) \( 2 x+y=7 \) 
\( \begin{array}{l}{\text { For } x=0} \\ {2(0)+y=7} \\ {\Rightarrow y=7}\end{array} \) 
Therefore, (0, 7) is a solution of this equation. 
\( \begin{array}{l}{\text { For } x=1} \\ {2(1)+y=7} \\ {\Rightarrow y=5}\end{array} \) 
Therefore, (1, 5) is a solution of this equation. 
\( \begin{array}{l}{\text { For } x=-1} \\ {2(-1)+y=7} \\ {\Rightarrow y=9}\end{array} \) 
Therefore, (-1, 9) is a solution of this equation. 
\( \begin{array}{l}{\text { For } x=2} \\ {2(2)+y=7} \\ {\Rightarrow y=3}\end{array} \) 
Therefore, (2, 3) is a solution of this equation. 
(ii) \( n x+y=9 \) 
\( \begin{array}{l}{\text { For } x=0} \\ {\pi(0)+y=9} \\ {\Rightarrow y=9} \\ {\text { Therefore, }(0,9) \text { is a solution of this equation. }}\end{array} \) 
\( \begin{array}{l}{\text { For } x=1} \\ {\pi(1)+y=9} \\ {\Rightarrow y=9-\pi} \\ {\text { Therefore, }(1,9-\pi) \text { is a solution of this equation. }}\end{array} \) 
\( \begin{array}{l}{\text { Therefore, }(2,9-2 \pi) \text { is a solution of this equation. }} \\ {\text { For } x=-1,} \\ {\pi(-1)+y=9} \\ {\Rightarrow y=9+\pi} \\ {\Rightarrow(-1,9+\pi) \text { is a solution of this equation. }}\end{array} \) 
(iii) \( x=4 y \) 
\( \begin{array}{l}{\text { For } x=0} \\ {0=4 y} \\ {\Rightarrow y=0} \\ {\text { Therefore, }(0,0) \text { is a solution of this equation. }}\end{array} \) 
\( \begin{array}{l}{\text { For } y=1} \\ {x=4(1)=4} \\ {\text { Therefore, }(4,1) \text { is a solution of this equation. }}\end{array} \) 
\( \begin{array}{l}{\text { For } y=-1} \\ {x=4(-1)} \\ {\Rightarrow x=-4} \\ {\text { Therefore, }(-4,-1) \text { is a solution of this equation. }}\end{array} \) 
\( \begin{array}{l}{\text { For } x=2} \\ {2=4 y} \\ {\quad y=\frac{2}{4}=\frac{1}{2}} \\ {\quad \quad \text { Therefore, }\left(2, \frac{1}{2}\right)_{\text { is a solution of this equation. }}}\end{array} \) 

Ans : (i) \( (0,2) \) 
\( \begin{array}{l}{\text { Putting } x=0 \text { and } y=2 \text { in the L.H.S of the given equation, }} \\ {x-2 y=0-2 \times 2=-4 \neq 4} \\ {\text { L.H. } 5 \neq \text { R.H.S }} \\ {\text { Therefore, }(0,2) \text { is not a solution of this equation. }}\end{array} \) 
(ii) \( (2,0) \) 
\( \begin{array}{l}{\text { Putting } x=2 \text { and } y=0 \text { in the L.H.S of the given equation, }} \\ {x-2 y=2-2 \times 0=2 \neq 4} \\ {\text { L.H. } 5 \neq \text { R.H.S }} \\ {\text { Therefore, }(2,0) \text { is not a solution of this equation. }}\end{array} \) 
(iii) \( (4,0) \) 
\( \begin{array}{l}{\text { Putting } x=4 \text { and } y=0 \text { in the L.H.S of the given equation, }} \\ {x-2 y=4-2(0)} \\ {=4=R . \text { H.S }} \\ {\text { Therefore, }(4,0) \text { is a solution of this equation. }}\end{array} \) 
(iv) \( (\sqrt{2}, 4 \sqrt{2}) \) 
\( \begin{array}{l}{\text { Putting } x=\sqrt{2} \text { and } y=4 \sqrt{2} \text { in the L.H.S of the given equation, }} \\ {x-2 y=\sqrt{2}-2(4 \sqrt{2})} \\ {\quad=\sqrt{2}-8 \sqrt{2}=-7 \sqrt{2} \neq 4} \\ {\text { L.H.S } \neq \mathrm{R.H.S}} \\ {\text { Therefore, }(\sqrt{2}, 4 \sqrt{2})_{\text { is not a solution of this equation. }}}\end{array} \) 
(v) \( (1,1) \) 
\( \begin{array}{l}{\text { Putting } x=1 \text { and } y=1 \text { in the L.H.S of the given equation, }} \\ {x-2 y=1-2(1)=1-2=-1 \neq 4} \\ {\text { L. H. S }  \neq \text { R.H.S }} \\ {\text { Therefore, }(1,1) \text { is not a solution of this equation. }}\end{array} \)