# Exercise E4.1 Linear Equations in Two Variables NCERT Solutions Class 9

## Chapter 4 Ex.4.1 Question 1

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

### Solution

#### Steps:

• Let the cost of one notebook be $$₹\, x$$
• Let the cost of one pen be $$₹\, y$$

Given: Cost of notebook is twice the cost of pen.

Therefore, we can write the required linear equation in two variables considering the given information as,

Cost of notebook $$= 2 \times$$ Cost of pen

$$\Rightarrow x = 2 y$$

Since any linear equation of two variables is expressed as:

$$a x + b y + c = 0,$$

$$\therefore \,x - 2 y = 0$$ is the required linear equation in two variables which represents the given information.

## Chapter 4 Ex.4.1 Question 2

Express the following linear equations in the form $$ax + by + c = 0$$ and indicate the values of $$a, b, c$$ in each case:

(i) \begin{align}2x + 3y = 9.3\overline 5\end{align}

(ii) \begin{align}x - \frac{y}abc - 10 = 0\end{align}

(iii) \begin{align}{-2 x+3 y}= Steps: (i) Consider \(\begin{align}&2x + 3y = 9.3\overline 5 \quad \dots {\rm{ Equation }}\left( {\rm{1}} \right)\\&\Rightarrow 2x + 3y - 9.3\overline 5 = 0\end{align}

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a=2{,}$$
• $$b=3{,}$$
• $$c=-9.3\overline {5}$$

(ii) Consider \begin{align}x-\frac{y}{5}-10=0 \quad \dots \text{Equation (1)}\end{align}

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a=1{,}$$
• \begin{align}b=-\frac{1}{5} \end{align}
• \begin{align}c=-10\end{align}

(iii) Consider

\begin{align}&-2 x+3 y=6 \quad \dots {\rm{ Equation }}\left( {\rm{1}} \right) \\ &{ \Rightarrow -2 x+3 y-6=0}\end{align}

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a=-2{,}$$
• $$a=-2{,}$$
• $$c=-6$$

(iv) Consider $$x = 3y \quad \dots \text{Equation (1)}$$

$$=> 1x - 3y + 0 = 0$$

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a=1{,}$$
• $$b=-3$$
• $$c=0$$

(v) Consider $$2x = -5y \qquad \dots \text{Equation (1)}$$

\begin{align}{ \Rightarrow 2x + 5y + 0 = 0}\end{align}

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a=2{,}$$
• $$b=5{,}$$
• $$c=0$$

(vi) Consider $$3x + 2 = 0 \quad \dots \text{Equation (1)}$$

We can re-write Equation ($$1$$) as shown below,

\begin{align}{ 3x + 0y + 2 = 0}\end{align}

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a=3{,}$$
• $$b=0{,}$$
• $$c=2$$

(vii) Consider $$\rm y - 2 = 0\quad \dots\text{Equation (1)}$$

We can re-write Equation ($$1$$) as shown below,

\begin{align}{ 0 x+1 y-2=0}\end{align}

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a = 0,$$
• $$b = 1,$$
• $$c = -2$$

(vii) $$5 = 2x \quad \dots \text{Equation (1)}$$

\begin{align}{ 2x + 0y - 5 =\rm 0}\end{align}

Comparing this equation with the standard form of the linear equation in two variables, $$ax + by + c = 0$$ we have,

• $$a = 2,$$
• $$b = 0,$$
• $$c = -5$$
\end{align}\)

(iv)  \begin{align}x={3 y}\end{align}

(v)  \begin{align}{2 x}={-\,5 y}\end{align}

(vi)  \begin{align}3 x+2={0}\end{align}

(vii) \begin{align}y-2={0}\end{align}

(viii)  \begin{align}5={2 x}\end{align}

### Solution

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