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

There are 4 exercises under NCERT Solutions for Class 9 Maths Chapter 4, which is Linear Equations in Two Variables. Students will get great practice by going through these solutions and will be able to master the concepts of Linear Equations in two Variables. Following are the details of question types and varieties that are included in each of the exercises:

• Exercise 4.1: Exercise 4.1 majorly includes questions related to determining the values of different variables and constructing linear equations in two variables to represent given statements.

• Exercise 4.2: Exercise 4.2 has problems comprising completing given statements and stating the reasons for choosing an answer, determining solutions for given equations, finding the actual solutions for given linear equations from provided solutions, and determining the values of constants.

• Exercise 4.3: Exercise 4.3 mostly has questions based on graphs. For example, plotting graphs, answering questions based on given graphs, providing solutions for given graphs, formulating relations and plotting relevant graphs for the same, choosing correct equations for provided graphs, deriving linear equations to satisfy given data and plotting the graph, are some of the questions that are included in this exercise.

• Exercise 4.4: Exercise 4.4 includes questions that are related to geometric representations, like giving or describing geometric representations as equations.

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - PDF Download

Exercise (4.1)

1: Construct a linear equation in two variables to express the following statement.

The cost of a textbook is twice the cost of an exercise book.

Ans. Let the cost of a textbook be $\text{x}$ rupees and the cost of an exercise book be $\text{y}$ rupees.

The given statement:  The cost of a textbook is twice the cost of an exercise book

So, in order to form a linear equation, 

the cost of the textbook $\text{=}\,\text{2 }\!\!\times\!\!\text{ }$ the cost of an exercise book.  

$\Rightarrow \text{x=2y}$

$\Rightarrow \text{x-2y=0}$.

2: Determine the values of $\text{a}$, $\text{b}$, $\text{c}$ from the following linear equations by expressing each of them in the standard form \[\text{ax+by+c=0}\].

(i) $\text{2x+3y=9}\text{.}\overline{\text{35}}$ 

Ans. The given linear equation is

$\text{2x+3y=9}\text{.}\overline{\text{35}}$

Subtracting $9.\overline{35}$ from both sides of the equation gives

$\text{2x+3y}-\text{9}\text{.}\overline{\text{35}}\text{=0}$ 

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as  

$\text{a=2}$, 

$\text{b=3}$, and

$\text{c}=-\text{9}\text{.}\overline{\text{35}}$

(ii) $\text{x-}\frac{\text{y}}{\text{5}}\text{-10=0}$

Ans. The given linear equation is

$\text{x-}\frac{\text{y}}{\text{5}}\text{-10}=\text{0}$ 

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{1}$,

$\text{b}=-\frac{\text{1}}{\text{5}}$, and

$\text{c}=-\text{10}$.

(iii) $\text{-2x+3y=6}$

Ans. The given linear equation is

$\text{-2x+3y=6}$

Subtracting $6$ from both sides of the equation gives 

$-\text{2x+3y}-\text{6}=\text{0}$

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{a}=-\text{2}$,

$\text{b}=\text{3}$, and

$\text{c}=-\text{6}$.

(iv) $\text{x=3y}$

Ans. The given linear equation can be written as

$\text{1x}=\text{3y}$

Subtracting $3y$ from both sides of the equation gives 

$\text{1x-3y+0=0}$

Now, by comparing the above equation with the standard form of the linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{a}=\text{1}$,

$\text{b}=-\text{3}$, and

$\text{c}=\text{0}$.

(v) \[\text{2x}\mathbf{=-}\,\text{5y}\]

Ans. The given linear equation is

\[\text{2x}=-\text{5y}\].

Adding $5y$ on both sides of the equation gives

\[\text{2x+5y+0=0}\].

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as 

$\text{a}=\text{2}$,

$\text{b}=\text{5}$, and

$\text{c}=\text{0}$.

(vi) $\text{3x+2=0}$

Ans. The given linear equation is

$\text{3x+2=0}$. 

Rewriting the equation gives

$\text{3x+0y+2=0}$

Now, by comparing the above equation with the standard form of linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{3}$,

$\text{b}=\text{0}$, and

$\text{c}=\text{2}$.

(vii) $\text{y-2=0}$

Ans. The given linear equation is

$\text{y-2=0}$ 

The equation can be expressed as

$\text{0x+1y-2}=\text{0}$

Now, by comparing the above equation with the standard form of the linear equation, $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=\text{0}$,

$\text{b}=\text{1}$, and

$\text{c}=-\text{2}$.

(viii) $\text{5=2x}$

Ans: The given linear equation is

$\text{5=2x}$.

The equation can be written as 

$\text{-2x+0y+5=0}$.

Now, by comparing the above equation with the standard form of the linear equation $\text{ax+by+c=0}$, the values of $\text{a,}\,\text{b,}$ and $\text{c}$ are obtained as

$\text{a}=-\text{2}$,

$\text{b}=\text{0}$, and

$\text{c}=\text{5}$.

Exercise (4.2) 

1: Complete the following statement by choosing the appropriate answer and explain why it should be chosen?

$\text{y=3x+5}$ has ___________. 

(a) A unique solution, 

(b) Only two solutions, 

(c) Infinitely many solutions. 

Ans: Observe that, $\text{y}=\text{3x+5}$ is a linear equation. 

Now, note that, for $\text{x}=\text{0}$, $\text{y}=\text{0+5=5}$.

So, $\left( \text{0,5} \right)$ is a solution of the given equation.

If $\text{x=1}$, then $\text{y}=\text{3 }\!\!\times\!\!\text{ 1+5}=\text{8}$.

That is, $\left( \text{1,8} \right)$ is another solution of the equation. 

Again, when $\text{y}=\text{0}$, $\text{x}=-\frac{5}{3}$ .

Therefore, $\left( -\frac{5}{3},0 \right)$ is another solution of the equation.

Thus, it is noticed that for different values of $\text{x}$ and $\text{y}$, different solutions are obtained for the given equation.

So, there are countless different solutions exist for the given linear equation in two variables. Therefore, a linear equation in two variables has infinitely many solutions. 

Hence, option (c) is the correct answer.

2: Determine any four solutions for each of equations given below. 

(i) $\mathbf{2x}+\mathbf{y}=\mathbf{7}$. 

Ans: The given equation 

$\text{2x+y}=\text{7}$ is a linear equation.

Solving the equation for $y$ gives

$\text{y=7-2x}$. 

Now substitute $\text{x=0,1,2,3}$ in succession into the above equation.

For  $\text{x=0}$, 

$\text{2}\left( \text{0} \right)\text{+y=7}$

$\Rightarrow \text{y=7}$

So, one of the solutions obtained is $\left( \text{x,y} \right)\text{=}\left( \text{0,7} \right)$. 

For  $\text{x=1}$, 

$\text{2}\left( \text{1} \right)\text{+y=7}$

$\Rightarrow \text{y=5}$

Therefore, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{1,5} \right)$.

For $\text{x=2}$, 

$\text{2}\left( \text{2} \right)\text{+y=7}$

$\Rightarrow \text{y=3}$

That is, a solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{3,1} \right)$.

Also, for $\text{x=3}$, 

\[\text{2}\left( \text{3} \right)\text{+y=7}\]

$\Rightarrow \text{y=1}$ 

So, another one solution is $\left( \text{x,y} \right)\text{=}\left( \text{3,1} \right)$.

Thus, four solutions obtained for the given equations are $\left( \text{0,7} \right)$ , $\left( 1,5 \right)$, $\left( 2,3 \right)$, $\left( 3,1 \right)$.

(ii) $\mathbf{\pi x}+\mathbf{y}=\mathbf{9}$.

Ans.The given equation 

$\pi x+y=9$                                                                          …… (a)

is a linear equation in two variables.

By transposing, the above equation (a) can be written as

$\text{y=9- }\!\!\pi\!\!\text{ x}$. 

Now substitute $\text{x=0,1,2,3}$ in succession into the above equation.

For $\text{x=0}$, 

$\text{y=9- }\!\!\pi\!\!\text{ }\left( \text{0} \right)$

$\Rightarrow \text{y=9}$

Therefore, one of the solutions obtained is $\left( \text{x,y} \right)\text{=}\left( \text{0,9} \right)$.

For $\text{x=1}$,

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{1} \right)$

$\Rightarrow \text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }$.

So, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{1,9- }\!\!\pi\!\!\text{ } \right)$.

For  $\text{x=2}$, 

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{2} \right)$

$\Rightarrow \text{y}=\text{9}-\text{2 }\!\!\pi\!\!\text{ }$

That is, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{2,}\,\text{9-2 }\!\!\pi\!\!\text{ } \right)$.

Also, for $\text{x=3}$, 

$\text{y}=\text{9}-\text{ }\!\!\pi\!\!\text{ }\left( \text{3} \right)$

$\Rightarrow \text{y}=9-\text{3 }\!\!\pi\!\!\text{ }$.

Therefore, another one solution is $\left( \text{x,y} \right)\text{=}\left( \text{3,}\,\text{9-3 }\!\!\pi\!\!\text{ } \right)$.

Thus, four solutions obtained for the given equations are $\left( 0,9 \right)$, $\left( \text{1,9,- }\!\!\pi\!\!\text{ } \right)$, $\left( \text{2,9-2 }\!\!\pi\!\!\text{ } \right)$, $\left( \text{3,9-3 }\!\!\pi\!\!\text{ } \right)$.

(iii) $\mathbf{x}=\mathbf{4y}$.

Ans. The given equation 

$\text{x=4y}$ is a linear equation in two variables.

By transposing, the above equation can be written as

$\text{y=}\frac{\text{x}}{4}$ .

Now substitute $\text{x=0,1,2,3}$ in succession into the above equation.

For $\text{x=0}$, 

$\text{y}=\frac{0}{4}=0$.

Therefore, one of the solutions is $\left( \text{x,y} \right)\text{=}\left( \text{0,0} \right)$.

For $x=1$,

$\text{y}=\frac{1}{4}$.

So, another solution of the given equation is $\left( \text{x,y} \right)\text{=}\left( \text{1,}\frac{1}{4} \right)$.

For $\text{x=2}$,

$\text{y}=\frac{2}{4}=\frac{1}{2}$.

That is, another solution obtained is $\left( \text{x,y} \right)\text{=}\left( \text{2,}\frac{1}{2} \right)$.

Also, for $\text{x=3}$.,

$\text{y}=\frac{3}{4}$.

Therefore, another one solution is $\left( \text{x,y} \right)=\left( 3,\frac{3}{4} \right)$.

Thus, four solutions obtained for the given equations are $\left( 0,0 \right)$, $\left( \text{1,}\frac{1}{4} \right)$, $\left( \text{2,}\frac{1}{2} \right)$, $\left( 3,\frac{3}{4} \right)$.

3: Identify the actual solutions of the linear equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] from each of the following solutions.

(i)  $\left( \mathbf{0},\mathbf{2} \right)$

Ans: Substituting $\text{x=0}$ and $\text{y=2}$ in the Left-hand-side of the equation \[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=0-2\left( 2 \right) \\ & =-4 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 0,2 \right)$.

Hence, $\left( 0,2 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

(ii) $\left( \mathbf{2},\mathbf{0} \right)$

Ans. Substituting $\text{x=2}$ and $\text{y=0}$ in the Left-hand-side of the equation \[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=2-2\left( 0 \right) \\ & =2 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 2,0 \right)$.

Hence, $\left( 2,0 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

(iii) $\left( \mathbf{4},\mathbf{0} \right)$ 

Ans. Substituting $\text{x=4}$ and $\text{y=0}$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=4-2\left( 0 \right) \\ & =4. \end{align}$

Therefore, Left-hand-side is equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 4,0 \right)$.

Hence, $\left( 4,0 \right)$ is a solution of the equation \[\text{x-2y=4}\]. 

(iv) $\left( \sqrt{\mathbf{2}}\mathbf{,4}\sqrt{\mathbf{2}} \right)$

Ans. Substituting $\text{x=}\sqrt{2}$ and $\text{y=4}\sqrt{2}$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=\sqrt{2}-2\left( 4\sqrt{2} \right) \\ & =\sqrt{2}-8\sqrt{2} \\ & =-7\sqrt{2} \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( \sqrt{2},4\sqrt{2} \right)$.

Hence, $\left( \sqrt{2},4\sqrt{2} \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

(v) $\left( \mathbf{1},\mathbf{1} \right)$

Ans. Substituting $\text{x}=1$ and $\text{y}=1$ in the Left-hand-side of the equation\[\text{ }\!\!~\!\!\text{ x-2y=4}\] gives

$\begin{align} & \text{x-2y}=1-2\left( 1 \right) \\ & =1-2 \\ & =-1 \\ & \ne 4. \end{align}$

Therefore, Left-hand-side is not equal Right-hand-side of the given equation for$\left( \text{x,y} \right)=\left( 1,1 \right)$.

Hence, $\left( 1,1 \right)$ is not a solution of the equation \[\text{x-2y=4}\]. 

4: If $\left( \mathbf{x},\mathbf{y} \right)=\left( \mathbf{2},\mathbf{1} \right)$ is a solution of the equation \[\text{2x+3y=k}\], then what is the value of $\mathbf{k}$?

Ans: By substituting $\text{x}=2$, $\text{y}=1$ and into the equation

\[\text{2x+3y=k}\] gives

$\text{2}\left( \text{2} \right)\text{+3}\left( \text{1} \right)\text{=k}$

$\Rightarrow \text{4+3=k}$

$\Rightarrow \text{k=7}$.

Hence, the value of $\text{k}$ is $7$.

Exercise (4.3)

1: Graph each of the linear equations given below.

(i) \[\text{ }\!\!~\!\!\text{ x+y=4}\]

Ans. The given linear equation is

\[\text{x+y=4}\]

\[\Rightarrow \text{y=4--x}\]                                                                                           …… (a)

Substitute $\text{x}=0$ into the equation (a) gives

$\text{y}=4-0=4.$

Similarly, substituting $\text{x}=2,4$ in succession into the equation (a), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,4 \right)$, $\left( 2,2 \right)$ and $\left( 4,0 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{x+y}=4$. 

(ii) \[\text{x--y=2}\]

Ans. The given linear equation is

\[\text{x-y}=2\]

\[\Rightarrow \text{y}=\text{x}-2\]                                                                                           …… (a)

Substitute $\text{x}=0$ into the equation (a) gives

$\text{y}=0-2=-2.$

Similarly, substituting $\text{x}=2,4$ in succession into the equation (a), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,-2 \right)$, $\left( 2,0 \right)$ and $\left( 4,2 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{x}-\text{y}=2$. 

(iii) \[\text{y=3x}\]

Ans. The given linear equation is

\[\text{y}=3\text{x}\]                                                                                                  …… (a)

Substitute $\text{x}=0$ into the equation (a) gives

$\text{y}=3\left( 0 \right)=0.$

Similarly, substituting $\text{x}=2,-2$ in succession into the equation (a), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,0 \right)$, $\left( 2,6 \right)$ and $\left( -2,-6 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{y}=3\text{x}$.

(iv) \[\text{3=2x+y}\]

Ans.  The given linear equation is

\[3=2\text{x+y}\]

$\Rightarrow \text{y}=3-2\text{x}$                                                                                      …… (a)

Substitute $\text{x}=0$ into the equation (a) gives

$\text{y}=3-2\left( 0 \right)=3$.

Similarly, substituting $\text{x}=1,\,3$ in succession into the equation (a), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,3 \right)$, $\left( 1,1 \right)$ and $\left( 3,-3 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $3=\text{2x}+\text{y}$.

2: Provided that the equations of two lines passing through the point \[\left( \mathbf{2,14} \right)\]. Can there exist more than two equations of such type? If it is, then state the reason. 

Ans. Provided that equations of two lines passing through \[\left( \text{2,14} \right)\]. 

It can be noted that the point \[\left( \text{2,14} \right)\] satisfies the equation \[\text{7x-y=0}\] and\[\text{x-y+12=0}\]. 

So, the equations \[\text{7x-y=0}\] and \[\text{x-y+12=0}\] represent two lines passing through a point \[\left( \text{2,14} \right)\]. 

Now, since we know that through infinite number of lines can pass through any one point, so, there are an infinite number such types of lines exist that passes through the point $\left( 2,14 \right)$. 

Hence, there exist more than two equations whose graph passes through the point $\left( 2,14 \right)$.

3: Determine the value of $\mathbf{a}$ in the linear equation \[\text{3y=ax+7}\] if the point \[\left( \mathbf{3,4} \right)\] lies on the graph of the equation.

Ans. Given that \[\text{3y=ax+7}\] is a linear equation and the point \[\left( 3,4 \right)\] lies on the equation.

Substituting $\text{x=3}$, \[\text{y=4}\] in the equation gives 

\[\text{3y=ax+7}\]

$\Rightarrow \text{3}\left( 4 \right)\text{=a}\left( 3 \right)\text{+7}$

$\Rightarrow \text{3a}=5$

$\Rightarrow \text{a}=\frac{5}{3}$.

Hence, the value of $\text{a}$ is $\frac{5}{3}$.

4: Derive a linear equation for the following situation: 

For the first kilometre, a cab take rent $\mathbf{8}$ rupees and for the subsequent distances it becomes $\mathbf{5}$ rupees per kilometres. Assume the distance covered is $\mathbf{x}$ km and total rent is $\mathbf{y}$ rupees. Hence, draw the graph of the linear equation.

Ans. Let the total distance covered $=$ $\text{x}$ km 

and the total cost for the distance travelled $=$$\text{y}$ rupees.

It is given that the rent for the 1st kilometre is $8$ rupees and for the subsequent km, it is $5$ rupees per kilometres.

Therefore, rent for the rest of the distance $=$ \[\left( \text{x-1} \right)\text{5}\] rupees.

Total cost for travelling $\text{x}$ km is given by

\[\text{y=}\left[ \text{8+}\left( \text{x-1} \right)\text{5} \right]\]

 \[\Rightarrow \text{y=8+5x-5}\]

 \[\Rightarrow \text{y=5x+3}\]                                                                                      …… (1)

\[\Rightarrow \text{5x-y+3=0}\],

which is the required linear equation.

Now, substituting $\text{x}=0$ into the equation (1) gives

$\text{y}=5\left( 0 \right)+3=3$.

Similarly, substituting $\text{x}=1,\,-1$ in succession into the equation (1), the following table of $\text{y}$ -values are obtained:

Now, Plot the points $\left( 0,3 \right)$, $\left( 1,8 \right)$ and $\left( -1,-2 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation \[\text{5x-y+3=0}\].

It is concluded by observing the graph of the linear equations that the variable $\text{x}$ and $\text{y}$ represent the distance travelled by the car and the total cost of rent for the distance respectively. Therefore, $\text{x}$ and $\text{y}$ are non-negative quantities.

Thus, only the first quadrant of the graph of the linear equation \[\text{5x-y+3=0}\] is only valid. 

5: Choose the correct linear equation for the given graphs in (a) and (b). 

(a) (i) $\mathbf{y}=\mathbf{x}$ 

     (ii) $\mathbf{x}+\mathbf{y}=\mathbf{0}$ 

     (iii) $\mathbf{y}=\mathbf{2x}$

     (iv) $\mathbf{2}+\mathbf{3y}=\mathbf{7x}$

Ans. It is observed in the given graph that the points $\left( -1,1 \right)$, $\left( 0,0 \right)$, and $\left( 1,-1 \right)$ lie on the straight line. Also, the coordinates of the points satisfy the equation \[\text{x+y=0}\].

So, \[\text{x+y=0}\] is the required linear equation corresponding to the given graph.

Hence, option (ii) is the correct answer. 

(b) (i) $\mathbf{y}=\mathbf{x}+\mathbf{2}$

     (ii) $\mathbf{y}=\mathbf{x}-\mathbf{2}$ 

     (iii) $\mathbf{y}=-\mathbf{x}+\mathbf{2}$

     (iv) $\mathbf{x}+\mathbf{2y}=\mathbf{6}$

Ans.It is observed in the given graph that the points $\left( -1,3 \right)$, $\left( 0,2 \right)$, and $\left( 2,0 \right)$ lie on the straight line. Also, the coordinates of the points satisfy the equation \[\text{y}=-\text{x+2}\].

So, \[\text{y}=-\text{x+2}\] is the required linear equation corresponding to the given graph.

Hence, option (iii) is the correct answer.

6: The work done by a body on the application of a constant force is proportional to the distance moved by the body. Formulate this relation by a linear equation and graph the same by using a constant force of five units. Hence from the graph, determine the work done when the distance moved by the body is 

(i) $\mathbf{2}$ units 

(ii) $\mathbf{0}$ unit. 

Ans: Let the distance moved by the body be $\text{x}$ units and the work done be $\text{y}$ units. 

Now, given that, work done is proportional to the distance. 

Therefore, \[\text{y}\propto \text{x}\].

\[\Rightarrow \text{y}=\text{kx}\],                                                                                        …… (a) 

where, $\text{k}$ is a constant.

By considering constant force of five units, the equation (a) becomes

$\text{y}=\text{5x}$.                                                                                             …… (b)

Now, substituting $\text{x}=0$ into the equation (b) gives

$\text{y}=5\left( 0 \right)=0$.

Similarly, substituting $\text{x}=1,-1$ in succession into the equation (b), gives the following table of $\text{y}$-values. 

Now, Plot the points $\left( 0,0 \right)$, $\left( 1,5 \right)$ and $\left( -1,-5 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{y}=\text{5x}$.

It can be concluded by observing the graph of the linear equation that the value of $\text{y}$ corresponding to \[\text{x=2}\] is $10$. Thus, when the distance moved by the body is $2$ units, then the work done by it is $10$ units. 

Also, the value of $\text{y}$ corresponding to \[\text{x=0}\] is $0$. So, when the distance travelled by the body is $0$ unit, then the work done by it is $0$ unit.

7: Derive a linear equation that satisfies the following data and graph it.

Sujata and Suhana, two students of Class X of a school, together donated $\mathbf{100}$ rupees to the Prime Minister’s Relief Fund for supporting the flood victims.

Ans: Let Sujata and Suhana donated $\text{x}$ rupees and $\text{y}$ rupees respectively to the Prime Minister’s Relief fund.  

Given that, the amount donated by Sujata and Suhana together is $100$ rupees.

Therefore, $\text{x+y}=100$. 

$\Rightarrow \text{y}=\text{100}-\text{x}$.                                                            …… (a)

Now, substituting $\text{x}=0$ into the equation (a) gives

$\text{y}=100-0=100$.

Similarly, substituting $\text{x}=50,100$ in succession into the equation (a), gives the following table of $\text{y}$-values. 

Now, Plot the points $\left( 0,100 \right)$, $\left( 50,50 \right)$ and $\left( 100,0 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{x+y}=100$.

It is concluded by observing the graph of the linear equation that the variable $\text{x}$ and $\text{y}$ are showing the amount donated by Sujata and Suhana respectively and so, $\text{x}$ and $\text{y}$ are nonnegative quantities. 

Hence, the values of $\text{x}$ and $\text{y}$ lying in the first quadrant will only be considered.

8: The following linear equation converts Fahrenheit to Celsius: 

$\mathbf{F=}\left( \frac{\mathbf{9}}{\mathbf{5}} \right)\mathbf{C+32}$,

where $\mathbf{F}$ denotes the measurement of temperature in Fahrenheit and $\mathbf{C}$ in Celsius unit.

Then do as directed in the following questions.

(i) Graph the linear equation given above by taking $\mathbf{x}$-axis as Celsius and $\mathbf{y}$-axis as Fahrenheit. 

Ans. The given linear equation is

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$                                                                                 …… (a) 

Now, substituting $C=0$ into the equation (a) gives

$\text{F}=\left( \frac{9}{5} \right)\left( 0 \right)+32=32$.

Similarly, substituting $C=-40,10$ in succession into the equation (a) gives the following table of $\text{F}$-values.

Now, Plot the points $\left( 0,32 \right)$, $\left( -40,-40 \right)$ and $\left( 10,50 \right)$ on a graph paper and connect the points by a straight line. 

Thus, the following graph of the straight line represents the required graph of the linear equation $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$.

(ii) Determine the temperature in Fahrenheit if it is $\mathbf{3}{{\mathbf{0}}^{\mathbf{o}}}\mathbf{C}$ in Celsius.

Ans. Given that the temperature $={{30}^{\circ }}\text{C}$.

Now, it is also provided that, $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$. 

Substitute $\text{C}=\text{32}$, in the above linear equation.

Then,

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{30+32=54+32=86}$.

Hence, the temperature in Fahrenheit obtained is \[\text{86  }\!\!{}^\circ\!\!\text{ F}\]. 

(iii) Determine the temperature in Celsius if it is \[\mathbf{9}{{\mathbf{5}}^{\mathbf{o}}}\mathbf{F}\] in Fahrenheit. 

Ans.The given temperature $=\text{9}{{\text{5}}^{\circ }}\text{F}$.

\[\text{F=?}\]

It is provided that, $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$  

Now, substitute $\text{F}=9\text{5}$, into the above linear equation.

Then it gives

$\text{95=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$

$\Rightarrow \text{63=}\left( \frac{\text{9}}{\text{5}} \right)\text{C}$

$\Rightarrow \text{C}=\text{35}$.

Hence, the temperature in Celsius obtained is \[\text{3}{{\text{5}}^{\circ }}\text{C}\]. 

(iv) Calculate the temperature in Fahrenheit when it is \[{{\mathbf{0}}^{\mathbf{o}}}\mathbf{C}\] in Celsius. Also, determine the temperature in Celsius when it is \[{{\mathbf{0}}^{\mathbf{o}}}\mathbf{F}\] in Fahrenheit.

Ans. It is known that, 

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{c+32}\text{.}$                                                                                  …… (a)

Now, substituting \[\text{C}=0\] in the above linear equation gives, 

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\left( 0 \right)\text{+32=32}$.

So, if \[\text{C}={{\text{0}}^{\text{o}}}\text{C}\], then \[\text{F}=\text{3}{{\text{2}}^{\circ }}\text{F}\].

Again, substituting  \[\text{F}=\text{0}\] into the equation (a) gives 

$\text{0=}\left( \frac{\text{9}}{\text{5}} \right)\text{C+32}$

$\Rightarrow \left( \frac{\text{9}}{\text{5}} \right)\text{C=}-\text{32}$

$\Rightarrow \text{C=}\frac{-\text{160}}{9}=-\text{17}\text{.77}$

Hence, if \[\text{F}={{\text{0}}^{\circ }}\text{F}\], then \[\text{C}=-\text{17}\text{.}{{\text{8}}^{\circ }}\text{C}\].

(v) Does there exist a temperature that numerically gives the same value in both Fahrenheit and Celsius? If it is, then show it. 

Ans:  It is provided that,

 $\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{c+32}$.

Let assume that \[\text{F=C}\].

Then, 

$\text{F=}\left( \frac{\text{9}}{\text{5}} \right)\text{F+32}$

$\Rightarrow \left( \frac{\text{9}}{\text{5}}-\text{1} \right)\text{F+32=0}$

$\Rightarrow \left( \frac{\text{4}}{\text{5}} \right)\text{F}=-\text{32}$

 $ \Rightarrow \text{F}=-\text{40}$.

Yes, there exists a temperature \[-40{}^\circ \] that gives numerically the same value in both Fahrenheit and Celsius.

Exercise (4.4)

1: Describe the geometric representation of \[\text{y=3}\] as an equation 

(i) in one variable 

Ans. The given equation is \[\text{y=3}\].

Note that, when \[\text{y=3}\] is considered as an equation in one variable, then actually it represents a number in the one-dimensional number line as shown in following figure.

(ii) in two variables.

Ans: The given equation is \[\text{y=3}\].

The above equation can be written as \[\text{0}\text{.x+y=0}\].

Note that when $\text{y=3}$ is considered in two variables, then it represents a straight line passing through point \[\left( 0,3 \right)\] and parallel to the $\text{x}$-axis. Therefore, all the points in the graph having the $\text{y}$-coordinate as $3$, contained in the collection. 

Hence, at \[\text{x=0}\], \[\text{y=3}\]; 

at \[\text{x=2}\], \[\text{y=3}\];  and

at \[\text{x}=-2\], \[\text{y=3}\] are the solutions for the given equation.

Now, Plot the points $\left( 0,3 \right)$, $\left( 2,3 \right)$ and $\left( -2,3 \right)$ on a graph paper and connect the points by a straight line. The graphical representation is shown below:

2: Give the geometric representations of \[\text{2x+9=0}\] as an equation 

(i) in one variable 

Ans. The given equation is \[\text{2x+9=0}\].

Now, the equation can be written as

\[\text{2x+9=0}\]

\[\Rightarrow \text{2x=(-9)}\]

$\Rightarrow \text{x=}\frac{\text{-9}}{2}=\text{-4}\text{.5}$

Hence, when \[\text{2x+9=0}\] is considered as an equation in one variable, then actually it represents a number $\text{x}=-4.5$ in the one-dimensional number line as shown in following figure 

(ii) in two variables 

Ans:  The given equation is \[\text{2x+9}=0\].

The above equation can be written as \[\text{2x+0y=}-\text{9}\].

Note that when \[\text{2x+9}=0\] is considered in two variables, then it represents a straight line passing through point \[\left( -4.5,0 \right)\] and parallel to the $\text{y}$-axis. Therefore, all the points in the graph having the $\text{x}$-coordinate as $-4.5$, contained in the collection. 

Hence, at \[\text{y=3}\], $\text{x}=-4.5$; 

at \[\text{y}=-1\], $\text{x}=-4.5$;  and

at \[\text{y}=1\], $\text{x}=-4.5$ are the solutions for the given equation.

Now, Plot the points $\left( -4.5,3 \right)$, $\left( -4.5,-1 \right)$ and $\left( -4.5,1 \right)$ on a graph paper and connect the points by a straight line. The graphical representation is shown below:

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - PDF Download

You can opt for Chapter 4 -  Linear Equations in Two Variables NCERT Solutions for Class 9 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

NCERT Solutions for Class 9 Maths

• Chapter 1 - Number System

• Chapter 2 - Polynomials 

• Chapter 3 - Coordinate Geometry

• Chapter 4 - Linear Equations in Two Variables

• Chapter 5 - Introductions to Euclids Geometry

• Chapter 6 - Lines and Angles

• Chapter 7 - Triangles

• Chapter 8 - Quadrilaterals

• Chapter 9 - Areas of Parallelogram and Triangles

• Chapter 10 - Circles

• Chapter 11 - Constructions

• Chapter 12 - Herons formula

• Chapter 13 - Surface area and Volumes

• Chapter 14 - Statistics

• Chapter 15 - Probability

Linear Equation in Two Variables

The Chapter 4 Linear Equation in Two Variables Class 9 is divided into five sections and four exercises. The first section is the introduction with no exercise. The Second and Third section discusses Linear Equation and it’s solution whereas the Fourth and Fifth sections are advanced topics where we learn about the graph of linear equations in two variables and the equations of lines parallel to x-axis and y-axis.

List of Exercises and topics covered in Linear Equation In Two Variable Class 9:

• Exercise 4.1 - Linear Equations

• Exercise 4.2 - Solution of a Linear Equation

• Exercise 4.3 - Graph of a Linear Equation in Two Variables

• Exercise 4.4 - Equations of Lines Parallel to the x-axis and y-axis

Equations Can Be Linear - Linear Equations

An equation includes equal sign (=) which indicates that the terms on the left-hand side are equal to the terms on the right-hand side. A Linear equation is an equation for a straight line containing variables and constants in the form given below:

\[a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}\]..... + b = 0

Where \[a_{1}, a_{2}, a_{3}\]... are coefficients, b is a constant and \[x_{1}, x_{2}, x_{3}\].... are the variables. If the value of any coefficient or variable is zero then the term containing that coefficient or variable becomes zero. This is because anything multiplied to zero is equal to zero.

A linear equation is a simple equation containing coefficients, constants and one or more variables, but a linear equation can never have exponents and roots.

One, Two, Three What Variables Would Be?- Types of Linear equations

One variable linear equation: A linear equation which has only one variable (unknown term) represented by alphabets or symbols is known as one variable linear equation. It is represented as ax+b = 0, where a is a coefficient of variable x and b is a constant. The coefficient can never be zero. 

Examples: 7x + 6 = 13

Two variable linear equation: A linear equation is an equation which has two variables (unknown terms) represented by alphabets or symbols known as a two-variable linear equation. It is represented as ax+by+c = 0, where a and b are coefficients,  x and y are variables and c is a constant. If any coefficient becomes zero then the two-variable linear equation changes to one variable linear equation.

Examples: 2x +3y = 24

Three or more variable linear equations: A linear equation containing more than two variables is called a linear equation of three or more variables. It can be represented as: 

\[a_{1}x_{1} + a_{2}x_{2} + a_{3}x_{3}\]..... + b = 0.

Examples: 5x+ 21y - 3z = -2

Linear Equation with Two Variables

Till now in the journey of algebraic equations, we have learned solving single equations with only one variable (unknown). For example something like 9x + 4 = 22. Simple, Isn’t it?

But what happens if there is more than one unknown in an equation something like 5x + 3y = 15. We solve it differently. So before looking into the solution of a linear equation with two variables let us understand the Linear equation with two variables mathematically.

Definition

An equation of the type ax+by+c = 0, where a,b,c are real numbers such that a and b are non-zero, is called a linear equation in two variables x and y.

Example: x+y-5 = 0 is a linear equation in the two variables(unknowns) x and y. Note that x=2 and y=3 satisfy this linear equation.

A Single Linear Equation with Two Variables Cannot be Solved.

There’s no way anyone could legitimately ask you to solve a single equation with two variables because that would give you infinite solutions. But two equations having two variables each can be solved to find the value of x and y simultaneously. A group of two or more equations is called a system of equations.

Each equation represents a straight line. If two lines are taken then there are high chances that those two lines intersect at a unique point which satisfies both the equations. In order to find the intersecting point, pick two random lines and solve.

Solution of a Linear Equation

We know that every linear equation in one variable has a unique solution. What about the solution of a linear equation with  two variables? There will always be a pair of values one for x and the other for y which satisfy the given equation. Also, note that there is no end to different solutions of a linear equation in two variables. That is, a linear equation in two variables has infinitely many solutions.

There are many ways to solve a system of linear equations with two variables. Given below are the two basic methods to solve a linear equation with two variables.

1. Graphical Method of Solving Linear Equation

Instead of finding the solution of two the linear equations separately we find the solution of the system instead. If we graph both the lines in the same coordinate system then the point of intersection of two lines will be the solution of the system.

For Example: To solve the system of equations having two equations -

2x+2 = y and x-1 = y, we need to consider a value of x and find its corresponding value of y for each equation. For equations 2x+2 = y and x-1 = y a random value of x is taken and its corresponding value of y is to be calculated. The points will be (1,4), (2,6), (3,8) for equation 2x+2 = y. And the points (1,0), (2,1), (3,2) for equation x-1 = y. The points are to be plotted on a graph. The point of intersection of these two lines will be the solution of the system.

2. Substitution Method of Solving Linear Equation

The other way of solving a system of linear equations is by substitution method. This system shows how to solve linear equations easily by finding the value of one variable in terms of another variable by using one equation and then replacing this value in another equation. 

Let's find the solution of the same system of linear equations.

2x+2 = y

x-1 = y

From equation (2) we can say that y = x-1.

Substituting the value of y in equation (1).

2x+2 = x-1

2x-x = -2-1

x = -3

Thus, we can find the actual value of y by substituting the value of x as -3 in equation (1).

2x + 2 = y

2(-3) + 2 = y

-6 + 2 = y

-4 = y

or, y= -4.

Therefore, the solution of the system of linear equations is (-3,-4).

NCERT Solutions Chapter- 4 Class 9 Maths By Vedantu

Vedantu provides learning in an entertaining and interesting manner to assist you in developing a firm conceptual foundation on each topic. Subject matter specialists give Class 9 Mathematics Chapter 4 Answers by tirelessly curating correct, simple, and step-by-step solutions for every question in NCERT textbooks. The numerical challenges are presented to assist you in approaching the chapter correctly and improving your comprehension of the key ideas. Mathematics Chapter 4 Solutions are created with the goal of covering the full curriculum in the form of NCERT answers. It has been shown to be a vital resource for students seeking effective learning in order to pass Board examinations and difficult competitive exams like as JEE (Mains and Advanced), AIMs, and others.

Vedantu being the best online tutoring company in India tries its best to render you real help by providing the NCERT Solutions for class 9th Maths Chapter 4Linear Equations and aim to deliver sufficient problems and solutions to practice and build a strong foundation on the chapter. The subject matter experts provide the NCERT Solutions for Class 9 Maths in a simple and precise manner with detailed summary provided at the end of the chapter. 

Deeper Into the Exercise - Types Of Questions In NCERT Class 9 Chapter 4

NCERT Grade 9 CBSE Chapter 4 Linear Equations in Two Variables belongs to Algebra. The Introduction describes solving a linear equation in two variables and how does the solution look like on the Cartesian plane. The topic Linear Equations explains about the points that should be kept in mind while solving a linear problem.In this chapter, you will get hold of the Solution of a Linear Equation. This topic explains a solution of a linear equation with two variables with a pair of values, one for x and one for y which satisfies the given equation.

All of these topics were presented through guided examples, making the learning process more participatory. The solutions to the issues between the chapters help students grasp their level of learning even more. This chapter will teach you interesting subjects like Graph of a Linear Equation in Two Variables and Equations of Lines Parallel to x-axis and y-axis by displaying the two variables of a linear equation on a graph sheet. The answers for all of these subjects are supplied step by step so that the student may internalise the notion gradually. Each lesson is followed by a short set of activities.

The exercises aim to test your knowledge and depth of understanding of the different theorems and concepts that are introduced in this chapter. Regardless, it must be noted that the numerical problems of this chapter are mostly based on specific theorems and other associated concepts. 

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables - PDF Download

All subjects are discussed through step-by-step solved examples in exercises. The solutions to the questions in this chapter will help you understand the fundamental notion of linear equations. A variety of solved examples of numerical problems are also provided to assist you enhance your comprehension of these topics and associated ideas. Furthermore, for each solved case, a detailed step-by-step explanation is provided. It can assist in understanding which strategies should be employed to approach various sorts of issues in order to solve them correctly. The Vedantu team validated the number of exercises and types of problems in class 9th mathematics chapter 4.

Section 1.2 - Exercise 1.1

The first exercise of this chapter consists of 2 questions with question number two having eight sub questions in exercise 1.2 of NCERT Solutions for Maths Class 9 Chapter 4. Most of the questions of this exercise are based on the standard form of linear equation with two variables which is a potent technique to compute the value of a,b and c of any given equation. There are basically three types of questions found from this section:

Type 1: Representing a statement in a linear equation with two variables.

Type 2: Expressing the linear equation in its standard form.

Type 3: Identification of a, b and c in a linear equation with two variables. 

These types of questions involve a lot of steps to reach the solution and hence comes with a risk of making a lot of silly mistakes. Make sure that you have a clear understanding of linear equations and variables. Also, a better understanding of the steps involved would help them clear their lingering doubts easily. Get all your doubts clear and strengthen your knowledge of the different concepts covered in the chapter by referring to our NCERT Solutions for Class 9th Maths Chapter 4. Each numerical problem has been explained step by step to make it easy for you to understand them and grasp the logic behind the same. Additionally, you will also find many helpful tips and alternative techniques to solve similar problems accurately and with more confidence.

Section 1.3 - Exercise 1.2

The second exercise in chapter 4 class 9 maths consists of 4 questions and is mostly based on the solution of linear equations with two variables. Once you grasp the concept of finding the solution, you will be able to identify equations having unique solutions, two solutions or many solutions.Maths class 9 chapter 4 has found wide application both in the field of mathematics and beyond. Given below are the types questions found related to the topic:

Type 1: Identification of the number of solutions of the given equations.

Type 2: Finding the solution of the given equation.

Type 3: Cross checking the solutions of the equation.

Type 4: Finding the value of an unknown term if the solution of the equation is given.

This exercise increases your knowledge in finding the solution of the equation. Doing so, you will gain more confidence as to how to find the unknown terms or how to identify the number of solutions of an equation. It will also prove useful in helping you solve similar types of numerical problems efficiently and in less time. Study the shortcut techniques from up close by taking a quick look at the NCERT Solutions for Class 9 Maths Chapter 4 pdf offered online in its PDF format.You can ace your upcoming board examination quite easily and with many conveniences by incorporating NCERT Solutions for Class 9 Maths Chapter 4 pdf into your revision plan. Download Vedantu’s study solutions from it’s learning portal with just a click and improve your learning experience without much ado.

Section 1.4 - Exercise 1.3

Third exercise of NCERT Solutions for Ch 4 Maths Class 9  consists of the maximum questions. The entire exercise is divided into eight questions, all of which are broadly based on the graph of linear equations with two variables.

Type 1: Drawing of the graph of a linear equation with two variables.

Type 2: Equation of a line passing through a point (x,y).

Type 3: Finding the value of the unknown term if the solution is given.

Type 4: Problem sums.

Type 5: Identification of equation from the graph.

Usually numerical problems involve lengthy steps and complex approaches, which is why it is vital to be well-versed with the fundamentals of the concepts they are based on.This exercise consists of important and complex of questions in Ch 4 Class 9 Maths Solutions so you grasp the fundamental concept of this section and start solving the questions effectively. Vedantu’s study guides like NCERT Solutions for Class 9 Maths Chapter 4 pdf have been engineered by the experts keeping in mind the needs and requirements of both the CBSE board examination and students.

Section 1.5 - Exercise 1.4

The last exercise of Chapter 4 Class 9 Maths consists of 2 questions with sub divisions and is mostly based on the concepts of geometrical representation of the equation of a line. The solution for Maths Class 9 Chapter 4 covers each of these concepts in depth to help students strengthen their grasp on the same. If you have a sound understanding of the topic then you will be able to apply the concept to solve different numerical problems. On the basis of theorems, the exercise can be segregated into the following question types:

Type 1: Geometrical representation of an equation in one variable and two-variable. 

It is advisable to solve the exercise and match your answer with our chapter-based solutions online, to gauge your understanding of the topics more effectively. Revising NCERT Solutions for Class 9th Maths Chapter 4 persistently will go a long way to help you ace your preparation for the upcoming board examination and will prove useful in scoring well in them. It will help you effectively solve this type of numerical problem accurately and in less time.

Download NCERT Solutions for Class 9 Maths from Vedantu, which are curated by master teachers. Also, you can revise and solve the important questions for class 9 Maths Exam 2023-24, using the updated NCERT Book Solutions provided by us. Subjects like Science, Maths, English will become easy to study if you have access to NCERT Solution for Class 9 Science, Maths solutions, and solutions of other subjects that are available on Vedantu only.

Importance of Class 9 Maths Chapter 4 Linear Equations in Two Variables NCERT Solutions

A linear equation of two variables is a function that two variables are linked. The power of the variables is one. This is why it is called a linear equation. Class 9 Maths Chapter 4 Linear Equations in Two Variables will explain how a linear equation can be plotted on a Cartesian coordinate plane for solutions. To understand the context of this chapter and the mathematical principles, refer to the NCERT solutions designed by the experts.

All the exercise questions are solved in a precise manner by following the CBSE Class 9 standards. These answers will enable students to learn and find out the context of the topics of this chapter. The solutions will also describe the ideal methods of decoding the requirement of a question and guide students to solve them in a stepwise manner.

Advantages of Class 9 Maths Chapter 4 Linear Equations in Two Variables NCERT Solutions

• Get the answers to all the exercise questions in one place. Find out how the experts have solved them in a precise way. Download the solution file or access it online to make your exercise solving sessions more productive.

• Use the solutions as the perfect material to develop your answering skills. Grab hold of the mathematical principles used in this chapter to solve problems related to linear equations with two variables.

• Resolve doubts related to the exercise questions on your own. Use the solutions to leave no queries unattended and take your preparation to the next level. Check how the experts have developed exclusive solving methods for particular questions and proceed accordingly.

Summary

• An equation of the form ax + by + c = 0, where a, b and c are real numbers, such that a and b are not both zero, is called a linear equation in two variables. 

• A linear equation in two variables has infinitely many solutions. 

• The graph of every linear equation in two variables is a straight line. 

• x = 0 is the equation of the y-axis and y = 0 is the equation of the x-axis. 

• The graph of x = a is a straight line parallel to the y-axis. 

• The graph of y = a is a straight line parallel to the x-axis. 

• An equation of the type y = mx represents a line passing through the origin. 

• Every point on the graph of a linear equation in two variables is a solution of the linear equation. Moreover, every solution of the linear equation is a point on the graph of the linear equation.