# NCERT Class 9 Maths Linear Equations in Two Variables

The chapter 4 starts with an introduction to linear equations followed by solutions to linear equations with few examples and some exercise questions. Later the chapter explains in detail how to draw the graph of linear equations in two variables followed by equations of a line parallel to x and y-axis.

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Linear Equations in Two Variables

Exercise 4.1

## 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**

**Video 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) \(\begin{align}x - \frac{y}{5} - 10 = 0\end{align}\)

(iii) \(\begin{align}{-2 x+3 y}={6}\end{align}\)

(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**

**Video Solution**

**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 *\(x = 3y \quad \dots \text{Equation (1)}\)*

\(=> 1x - 3y + 0 = 0 \)

- \(a=1{,}\)
- \(b=-3\)
- \(c=0\)

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

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

- \(a=2{,}\)
- \(b=5{,}\)
- \(c=0\)

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

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

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

- \(a=3{,}\)
- \(b=0{,}\)
- \(c=2\)

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

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

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

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

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

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

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