# Ex.4.2 Q3 Linear Equations in Two Variables Solution - NCERT Maths Class 9

## Question

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) (\(1, 1\))

## Text 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.