Linear Equations in Two Variables – NCERT Solutions Class 9

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables is an important topic for Class 9 and students need to prepare it wisely to score good marks in their examination. The chapter holds significant weightage, thus, it must be taken seriously by students.

If students are facing problems in the chapter or have a few doubts, SimplyAcad has provided solutions to all the exercises from the chapter in a detailed way. Scroll below to get all the answers explained structurally with their exercise and question numbers mentioned.

Overview of the Exercises Of NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations In Two Variables

• Exercise 4.1: The first exercise deals with the standard form of linear equation having two variables. The questions are based on representing a statement, expressing the linear equation in two variables form, and then identifying a,b,c in a linear equation.
• Exercise 4.2: The second exercise is the sub part of the first and mainly focuses on the above mentioned two variable linear equations. Identifying the number of solutions of the equation, finding the solution and the unknown term is the basis of the questions asked here.
• Exercise 4.3: The third exercise deals with drawing graphs of the given data. This section also includes a few word problems to be solved.
• Exercise 4.4 : The final exercise includes geometrical representations of the given set of data.

Linear Equations In Two Variables Class 9: NCERT Solutions for Chapter 4 Exercise 4.1

Question 1: The cost of a notebook is twice the cost of a pen. Write a linear equation in two variables to represent this statement.

Solution: Let the cost of the pen be y and the cost of notebook be x.

According to statement,

Cost of a notebook = twice the pen = 2y.

\(\therefore 2y = x\)

\(\Rightarrow x – 2y = 0\)

This is a linear equation in two variables to represent this statement.

Question 2: Express the following linear equations in the form $a x+b y+c=0$ 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) $5=2 x$

Solution:

\((i) 2x + 3y = 9.35\)

\(\Rightarrow 2x + 3y – 9.35 = 0\)

On comparing this equation with \(ax + by + c = 0,\) we get,

\(a = 2, b = 3~ and ~c = -9.35\)

\((ii) x – \frac y5 – 10 = 0\)

On comparing this equation with \(ax + by + c = 0\), we get,

\(a = 1, b = \frac{-1}{5} ~and~ c = -10\)

\((iii) -2x + 3y = 6\)

\(\Rightarrow -2x + 3y – 6 = 0\)

On comparing this equation with \(ax + by + c = 0\), we get,

\(a = -2, b = 3~ and~ c = -6\)

\((iv) x = 3y\)

\(\Rightarrow x – 3y = 0\)

On comparing this equation with \(ax + by + c = 0,\) we get,

\(a = 1, b = -3 ~and~ c = 0\)

\((v) 2x = -5y\)

\(\Rightarrow 2x + 5y = 0\)

On comparing this equation with \(ax + by + c = 0\), we get,

\(a = 2, b = 5 ~and~ c = 0\)

\((vi) 3x + 2 = 0\)

\(\Rightarrow 3x + 0y + 2 = 0\)

On comparing this equation with \(ax + by + c = 0,\) we get,

\(a = 3, b = 0 ~and~ c = 2\)

\((vii) y – 2 = 0\)

\(\Rightarrow 0x + y – 2 = 0\)

On comparing this equation with \(ax + by + c = 0\), we get,

\(a = 0, b = 1~ and~ c = -2\)

\((viii) 5 = 2x\)

\(\Rightarrow -2x + 0y + 5 = 0\)

On comparing this equation with \(ax + by + c = 0,\) we get,

\(a = -2, b = 0 ~and~ c = 5\)

Linear Equations In Two Variables Class 9: NCERT Solutions for Chapter 4 Exercise 4.2

Question 1: 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

Solution:

Since the equation, \(y = 3x + 5\) is a linear equation in two variables, it will have infinitely many solutions.

Different values of \(x\) gives corresponding values of \(y\).

Question 2: Write four solutions for each of the following equations:

\((i) 2x + y = 7 ~ (ii) \pi x + y = 9 ~ (iii) x = 4y\)

Solution:

\((i) 2x + y = 7\)

\(\Rightarrow y = 7 – 2x\)

\(\to ~Put~ x = 0,\)

\(y = 7 – 2 \times 0 \Rightarrow y = 7\)

\((0, 7)\) is a solution.

\(\to~ Now, ~put ~x = 1\)

\(y = 7 – 2 \times 1 \Rightarrow y = 5\)

\((1, 5)\) is a solution.

\(\to~ Now,~ put~ x = 2\)

\(y = 7 – 2 \times 2 \Rightarrow y = 3\)

\((2, 3)\) is a solution.

\(\to~ Now,~ put~ x = -1\)

\(y = 7 – 2 \times -1 \Rightarrow y = 9\)

\((-1, 9)\) is a solution.

\((ii) \pi x + y = 9\)

\(\Rightarrow y = 9 – \pi x\)

\(\to~ Put~ x = 0,\)

\(y = 9 – \pi \times0 \Rightarrow y = 9\)

\((0, 9)\) is a solution.

\(\to~ Now, ~put~ x = 1\)

\(y = 9 – \pi \times1 \Rightarrow y = 9-\pi\)

\((1, 9-\pi )\) is a solution.

\(\to~ Now, ~put ~x = 2\)

\(y = 9 – \pi\times2 \Rightarrow y = 9-2\pi\)

\((2, 9-2\pi)\) is a solution.

\(\to~ Now, ~put~ x~ = -1\)

\(y = 9 – \pi \times -1 \Rightarrow y = 9+\pi\)

\((-1, 9+\pi)\) is a solution.

\((iii) x = 4y\)

\(\to~ Put~ x = 0,\)

\(0 = 4y\)

\( \Rightarrow y = 0\)

\((0, 0)\) is a solution.

\(\to~ Now,~ put~ y = 1\)

\(x = 4(1) \Rightarrow 4\)

\((4, 1)\) is a solution.

\(\to~ Now,~ put~ y = -1\)

\(x = 4(-1)\)

\(x = -4\)

Therefore, \((-4, -1)\) is a solution of this equation.

For \(x = 2,\)

\(2 = 4y\)

\(y = \frac24 = \frac12\)

Therefore, \((2, \frac12)\) is a solution of this equation.

Question 3: Check which of the following are solutions of the equation \(x – 2y = 4\) and which are not:

\((i) ~(0, 2) ~~~(ii)~ (2, 0)~~~ (iii) ~(4, 0)~~~ (iv)~ (\surd2, 4\surd2)~~~ (v)~ (1, 1)\)

Solution:

(i) Put \(x = 0 ~and ~y = 2\) in the equation \(x – 2y = 4.\)

\(0 – 2\times 2 = -4\)

\(\Rightarrow -4 \neq 4\)

\(\therefore (0, 2)\) is not a solution of the given equation.

(ii) Put \(x = 2 ~and~ y = 0\) in the equation \(x – 2y = 4.\)

\(2 – 2\times0 = 2\)

\(\Rightarrow 2 \neq 4\)

\(\therefore (2, 0)\) is not a solution of the given equation.

(iii) Put \(x = 4 ~and~ y = 0\) in the equation \(x – 2y = 4.\)

\(4 – 2\times0 = 4\)

\(\Rightarrow 4 = 4\)

\(\therefore (4, 0)\) is a solution of the given equation.

\((iv) ~Put~ x = \surd2 ~and~ y = 4\surd2\) in the L.H.S equation \(x – 2y = 4.\)

\(\surd2 – 2\times4\surd2 = \surd2 – 8\surd2 = \surd2(1 – 8) = -7\surd2\)

\(\Rightarrow -7\surd2 \neq 4\)

\(\therefore (\surd2, 4\surd2)\) is not a solution of the given equation.

(v) Put \(x = 1~ and~ y = 1\) in the equation \(x – 2y = 4.\)

\(1 – 2\times1 = -1\)

\(\Rightarrow -1 \neq 4\)

\(\therefore (1, 1)\) is not a solution of the given equation.

Question 4: Find the value of k, if \(x = 2, y = 1\) is a solution of the equation \(2x + 3y = k.\)

Solution:

\(= 2x + 3y = k\)

Given

\(x = 2, y = 1\) is the solution of the given equation.

Putting the value of x and y in the equation \(2x + 3y = k\), we get,

\(2\times2 + 3\times1 = k\)

\(\Rightarrow k = 4 + 3\)

\(\Rightarrow k = 7 \)