**NCERT Solutions Class 9 Maths Chapter 4 Linear Equations In Two Variables** – Here are all the NCERT solutions for Class 9 Maths Chapter 4. This solution contains questions, answers, images, explanations of the complete Chapter 4 titled Linear Equations In Two Variables of Maths taught in class 9. If you are a student of class 9 who is using NCERT Textbook to study Maths, then you must come across Chapter 4 Linear Equations In Two Variables. After you have studied lesson, you must be looking for answers of its questions. Here you can get complete NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables in one place.

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

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Class | 9 |

Subject | Maths |

Book | Mathematics |

Chapter Number | 4 |

Chapter Name |
Linear Equations In Two Variables |

### NCERT Solutions Class 9 Maths chapter 4 Linear Equations In Two Variables

Class 9, Maths chapter 4, Linear Equations In Two Variables solutions are given below in PDF format. You can view them online or download PDF file for future use.

### Linear Equations In Two Variables

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### Question & Answer

Q.1:The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement. (Take the cost of a notebook to be ₹ x and that of a pen to be ₹ y).

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

Q.2:Express the following linear equations in the form ax + by + c = 0 and indicate the values of a, b and c in each case: (i) \( 2 x+3 y=9.3 \overline{5} \) (ii) \( x-\frac{y}{5}-10=0 \) (iii) \( -2 x+3 y=6 \) (iv) \( x=3 y \) (v) \( 2 x=-5 y \) (vi) \( 3 x+2=0 \) (vii) \( y-2=0 \) (viii) \( 5=2 x \)

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) \( 5=2 x \) \( \begin{array}{l}{-2 x+0 . y+5=0} \\ {\text { Comparing this equation with } a x+b y+c=0} \\ {a=-2, b=0, c=5}\end{array} \)

Q.3:Which one of the following options is true, and why? y = 3x + 5 has (i) a unique solution, (ii) only two solutions, (iii) ) infinitely many solutions

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).

Q.4:Write four solutions for each of the following equations: (i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

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) \( x=4 y \) \( \begin{array}{l}{\text { For } x=0} \\ {0=4 y} \\ {\Rightarrow y=0} \\ {\text { Therefore, }(0,0) \text { is a solution of this equation. }}\end{array} \) \( \begin{array}{l}{\text { For } y=1} \\ {x=4(1)=4} \\ {\text { Therefore, }(4,1) \text { is a solution of this equation. }}\end{array} \) \( \begin{array}{l}{\text { For } y=-1} \\ {x=4(-1)} \\ {\Rightarrow x=-4} \\ {\text { Therefore, }(-4,-1) \text { is a solution of this equation. }}\end{array} \) \( \begin{array}{l}{\text { For } x=2} \\ {2=4 y} \\ {\quad y=\frac{2}{4}=\frac{1}{2}} \\ {\quad \quad \text { Therefore, }\left(2, \frac{1}{2}\right)_{\text { is a solution of this equation. }}}\end{array} \)

Q.5:Check which of the following are solutions of the equation x – 2y = 4 and which are not: (i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) \( (\sqrt{2}, 4 \sqrt{2}) \) (v) (1, 1)

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) \( (1,1) \) \( \begin{array}{l}{\text { Putting } x=1 \text { and } y=1 \text { in the L.H.S of the given equation, }} \\ {x-2 y=1-2(1)=1-2=-1 \neq 4} \\ {\text { L. H. S } \neq \text { R.H.S }} \\ {\text { Therefore, }(1,1) \text { is not a solution of this equation. }}\end{array} \)

## NCERT / CBSE Book for Class 9 Maths

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### All NCERT Solutions Class 9

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