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

## 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**

**Video 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**

**Video 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**

**Video 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 \(7.\)