# Exercise E4.2 Linear Equations in Two Variables NCERT Solutions Class 9

Go back to  'Linear Equations in Two Variables'

## Chapter 4 Ex.4.2 Question 1

Which one of the following options is true, and why?

$$y = 3x + 5$$ has

(i) A unique solution,

(ii) only two solutions,

(iii) infinitely many solutions

#### ### Solution Video Solution

What is known?
Linear equation \begin{align} y = 3x + 5\end{align}

What is Unknown?
Number of solutions of the given equation.

Reasoning:
We can check number of solution by putting different values of $$x$$ and get different values of $$y$$.

Steps:

We know that

• $$y = 3x + 5$$ is a linear equation in two variables in the form of $$ax + by + c = 0$$
• For $$x = 0, y = 0 + 5 = 5$$          Therefore, $$(0, 5)$$ is one solution.
• For $$x = 1, y = 3×1 + 5 = 8$$      Therefore, $$(1, 8)$$ is another solution.
• For $$y = 0, 3x + 5 = 0, x = - 5/3$$   Therefore, $$(- 5/3 , 0)$$ is another solution .

Clearly, for different values of $$x$$, we get various values for $$y$$.

Thus, any value substituted for $$x$$ in the given equation will constitute another solution for the given equation.

So, there is no end to the number of different solutions obtained on substituting real values for $$x$$ in the given linear equation. Therefore, a linear equation in two variables has infinitely many solutions.

Hence (iii) is the correct answer

## Chapter 4 Ex.4.2 Question 2

Write four solutions for each of the following equations:

(i) $$2 \mathrm{x}+\mathrm{y}=7$$

(ii) $$\pi x+y=9$$

(iii) $$x=4 y$$

### Solution

What is known?
Linear equations

What is Unknown?
Four solutions of each equation.

Reasoning:
We can find any number of solutions by putting different values of $$x$$ and get different values of $$y$$.

Steps:

(i) $$2x + y = 7$$

Given,

Linear Equation, $$2x + y = 7 \qquad \dots \dots \text{Equation (1)}$$

$$\therefore y = 7- 2x\qquad \dots \dots \text{Equation (1)}$$

Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get

For $$x = 0$$, we get $$y = 7- 2(0)\,\Rightarrow y = 7$$. Hence, we get $$(x, y) = (0, 7)$$

For $$x = 1$$, we get $$y = 7- 2(1)\,\Rightarrow y = 5$$. Hence, we get $$(x, y) = (1, 5)$$

For $$x = 2$$, we get $$y = 7- 2(2) \,\Rightarrow y = 3$$. Hence, we get $$(x, y) = (2, 3)$$

For $$x = 3$$, we get $$y = 7- 2(3) \,\Rightarrow y = 1$$. Hence, we get $$(x, y) = (3, 1)$$

Therefore, the four solutions of the given equation are ($$0,7$$), ($$1,5$$), ($$2, 3$$) and ($$3,1$$).

(ii) $$\quad \pi x+y=9$$

Given,

Linear Equation, $$\pi x+y=9$$

$$y=9-\pi x \qquad \dots \dots \text{Equation (1)}$$

Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get

For \begin{align},x = 0,y = 9 - \pi \left( 0 \right) \Rightarrow y=9.\end{align}
Hence, we get \begin{align}\left( {x,y} \right) = (0,9 )\end{align}
For \begin{align},x = 1,y = 9 - \pi \left( 1 \right) \Rightarrow 9 - \pi .\end{align}
Hence, we get \begin{align}\left( {x,y} \right) = (1,9 - \pi )\end{align}
For \begin{align},x = 2,y = 9 - \pi \left( 2 \right) \Rightarrow 9 - 2\pi .\end{align}
Hence, we get \begin{align}\left( {x,y} \right) = (2,9 - 2\pi )\end{align}
For \begin{align},x = 3,y = 9 - \pi \left( 3 \right) \Rightarrow 9 - 3\pi .\end{align}
Hence, we get \begin{align}\left( {x,y} \right) = (3,9 - 3\pi )\end{align}

Therefore, the four solutions of the given equation are $$(0,9),(1,9-\pi),(2,9-2 \pi),(3,9-3 \pi)$$.

(iii) $$x = 4y$$

Given,

Linear Equation, $$x = 4y$$

\begin{align}\therefore y = \frac{x}{4} \qquad \dots \dots \text{Equation (1)}\end{align}

Let us now take different values of $$x$$ and substitute in Equation ($$1$$), we get

For $$x = 0$$, \begin{align}y =\frac{0}{4} = 0\end{align}.

Hence, we get $$(x, y)$$ $$= (0, 0)$$

For $$x = 1$$, \begin{align}y =\frac{1}{4}\end{align}

Hence, we get \begin{align}(x, y) = (1,\frac{1}{4} )\end{align}

For $$x = 2$$, \begin{align}y =\frac{2}{4} =\frac{1}{2}\end{align}

Hence, we get \begin{align}(x, y) = (2,\frac{1}{2} )\end{align}

For $$x = 3$$,\begin{align}y = \frac{3}{4}\end{align}.

Hence, we get \begin{align}(x, y) = (3,\frac{3}{4} )\end{align}

Therefore, the four solutions of the given equation are \begin{align}(0, 0), (1,\frac{1}{4}), (2, \frac{1}{2})\end{align} and \begin{align}(3,\frac{3}{4})\end{align}

## Chapter 4 Ex.4.2 Question 3

Check which of the following solutions of the equation are $$x - 2y = 4$$ and which are not:

(i) ($$0, 2$$)

(ii) ($$2, 0$$)

(iii) ($$4, 0$$)

(iv) ( $$\sqrt 2 , \,4 \sqrt2$$)

(v) ($$1, 1$$)

### Solution

What is known?
Linear equation $$x − 2y = 4$$

What is Unknown?
Given values are solution or not of the equation.

Reasoning:
We can substitute the values in the given equation and can check whether $$LHS$$ is equal to $$RHS$$ or not.

Steps:

#### Given:

$$x - 2y = 4$$ is a Linear Equation of the form $$ax + by + c = 0 \quad \dots \text{Equation(1)}$$

(i) Consider ($$0, 2$$)

By Substituting $$x = 0$$ and $$y = 2$$ in the given Equation ($$1$$)

\begin{align}x - 2y &= 4\\{\rm{0 }}-{\rm{ }}2\left( 2 \right) &= 4\\0 - 4 &= 4\\{\rm{ - 4 }} &\ne {\rm{ }}4\\\\L.H.S{\rm{ }} &\ne {\rm{ }}R.H.S\end{align}

Therefore, ($$0, 2$$) is not a solution of this equation.

(ii) Consider ($$2, 0$$)

By Substituting, $$x = 2$$ and $$y = 0$$ in the given Equation ($$1$$),

\begin{align}x - 2y&= 4\\2{\rm{ }}-{\rm{ }}2\left( 0 \right) &= 4\\2 - 0 &= 4\\2{\rm{ }} &\ne {\rm{ }}4\\\\L.H.S{\rm{ }} &\ne {\rm{ }}R.H.S\end{align}

Therefore, ($$2, 0$$) is not a solution of this equation

(iii) Consider ($$4, 0$$)

By Substituting, $$x = 4$$ and $$y = 0$$ in the given Equation ($$1$$)

\begin{align}x - 2y &= 4\\{4{\rm{ }} - \rm{ }}2\left( 0 \right) &= 4\\4 - 0 &= {\rm{ }}4\\4 &= 4\\\\L.H.S{\rm{ }} &= {\rm{ }}R.H.S\\\end{align}

Therefore, ($$4, 0$$) is a solution of this equation.

(iv) $$\sqrt{2}, 4 \sqrt{2}$$

By Substituting, $$x=\sqrt{2} \text { and } y=4 \sqrt{2}$$ in the given Equation ($$1$$)

\begin{align}x - 2y &= 4\\\sqrt 2 - 8\sqrt 2 &= 4\\ - 7\sqrt 2 & \ne {\rm{ }}4\\\\{\text{Thus, }}L.H.S &\ne R.H.S\end{align}

Therefore,\begin{align}(\sqrt 2 ,4\sqrt 2 )\end{align} is not a solution of this equation.

(v) ($$1, 1$$)

By Substituting, $$x = 1$$ and $$y = 1$$ in the given Equation ($$1$$)

\begin{align}x{\rm{ }} - {\rm{ }}2y &= 4\\1{\rm{ }} - {\rm{ }}2\left( 1 \right) &= 4\\1{\rm{ }} - {\rm{ }}2 &= 4\\ - 1{\rm{ }}& \ne {\rm{ }}4\\\\{\text{Thus, }}L.H.S{\rm{ }} &\ne {\rm{ }}R.H.S\end{align}

Therefore, ($$1,1$$) is not a solution of this equation.

## Chapter 4 Ex.4.2 Question 4

Find the value of $$k,$$ if $$x = 2, y = 1$$ is a solution of the equation $$2x + 3y = k.$$

### Solution

What is known?

Linear equation \begin{align}2x + 3y = k.\end{align}

What is Unknown?

Value of $$K$$.

Reasoning:

We can find value of $$k$$ by substituting the values of $$x$$ and $$y$$ in the given equation.

#### Steps:

Given :

$$2x + 3y = k$$ is the Linear Equation ---------------- Equation ($$1$$)

• $$x = 2$$
• $$y = 1$$
• $$k =?$$

By substituting the values of $$x$$ and $$y$$ in Equation ($$1$$),

\begin{align}&2x + 3y = k\\&\Rightarrow(2)(2) + 3(1) = k\\&\Rightarrow4 + 3 = k\end{align}

Hence, $$k = 7$$

Therefore, the value of $$k$$ is $$7.$$