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

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 $\text{x+y}=4$. 

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(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 $\text{x}-\text{y}=2$. 

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(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 $\text{y}=3\text{x}$.

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(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 $3=\text{2x}+\text{y}$.

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

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

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

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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 $\text{y}=\text{5x}$.

(Image will be Uploaded Soon)

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 $\text{x+y}=100$.

(Image will be Uploaded Soon)

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

(Image will be Uploaded Soon)

(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.

(Image will be Uploaded Soon)

(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:

(Image will be Uploaded Soon)

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 

(Image will be Uploaded Soon)

(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:

(Image will be Uploaded Soon)

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).

Why NCERT Solutions For Class 9 Maths?

Every exam has a pattern to test students' knowledge and their ability to perform in the near future. Solving NCERT Solutions of Linear Equations In Two Variables Class 9 after studying the chapter helps you to understand the exam pattern, weightage of each topic, and furnishes your exam preparation. Practising Class 9 Maths Chapter 4 Solutions increases your speed during exams because a lot of students fail to complete the paper on time. Practising drills in you the tendency to think of the solution of the given question at breakneck speed. Also, you can predict the topics that can come in the exam by gauging the frequency and interval of the repeated questions. This can help you strategize your preparation and score good marks. With practice, you can also solve the unrepeated or brand new questions too because the concept of questions in the question papers usually remains the same. It’s just the twisted questions and changed numbers that brings the difference. You also get used to the level or standard of board exam questions. So understand and know the concept while solving questions and you will end up scoring above 90 per cent. It is a well known fact that some of us prepare to score maximum but end up getting less because of our carelessness or conceptual mistakes. And this can be corrected by a proper practice. Practicing maths problems not just enhances your problem-solving skills but also amplifies your analytical and reasoning abilities which is much needed in today's generation. It is important for all of us to understand and identify problems, think logically, interpret, make decisions and solve the problem. 

NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables Free PDF available on CoolGyan are solved by experts. Download Free Study Material for Class 9 to score more marks.

NCERT Solutions Chapter- 4 Class 9 Maths By CoolGyan

CoolGyan offers learning in an enjoyable and interesting way to help you develop a strong conceptual base on every chapter. Class 9 Maths Chapter 4 Solutions are provided by subject matter experts who untiringly work together in curating the accurate, easy and step-by-step solutions for every question in NCERT textbooks. The numerical problems are provided to help you attain the right approach to the chapter and improve your understanding of the important concepts. Maths Chapter 4 Solutions are prepared with an aim to cover the entire syllabus in the form of solutions to the NCERT questions. It is proven to be an important material for students looking for effective learning to crack the Board exams and tough competitive exams like JEE(Mains and Advanced), AIMs, etc. 

CoolGyan 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. 

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

Key Benefits Of NCERT Class 9 Maths Chapter 4 Solutions By CoolGyan:

• The solutions are written in simple language keeping in mind the average age group of the students.

• Complicated solutions are broken down into simple steps to make it easy for you to grasp the concept in less time.

• Basic facts, terms, principles and applications are highlighted.

• The solutions are prepared strictly as per latest NCERT solutions.

• It covers the entire syllabus and concept in the form of solutions.

• The answers are treated systematically and in an interesting manner.

• The content is kept concise, brief and self-explanatory.

• Some answers include necessary images to facilitate the understanding of the concept.

• It is handy and serves as a note during exam revision.

• The solutions are kept easy for you to solve maximum questions and get a command of the chapter. Also, it Improves the problem solving speed

Deeper Into the Exercise - Types Of Questions

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 concepts have been taught with the help of guided examples, thus making the learning process more interactive. The solutions pertaining to the problems between the chapters further help the students to understand their level of learning. Interesting topics such as Graph of a Linear Equation in Two Variables and Equations of Lines Parallel to x-axis and y-axis will be learnt in this chapter with the help of plotting the two variables of a linear equation on a graph sheet. The solutions provided for all these topics are step by step so that the learner can imbibe the concept step by step. Each of the topics is followed by compact exercises. 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 the topics are explained with the help of step by step solved examples in exercises. The solutions given for the problems in this topic will help you to become aware of the basic concept of linear equations. To further help you improve your understanding of these topics and related concepts, numbers of solved examples of numerical problems are also offered. Moreover, a thorough step by step explanation is provided for each solved example. It can help understand which methods are to be used to approach different types of questions for solving them accurately. The CoolGyan team has verified how many exercises and types of questions are there in class 9th maths 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 CoolGyan’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. CoolGyan’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 CoolGyan, which are curated by master teachers. Also, you can revise and solve the important questions for class 9 Maths Exam 2019 - 2020, 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 CoolGyan only.

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.