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

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

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

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,

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