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

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## 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) ($$1, 1$$)

Video Solution
Linear Equations In Two Variables
Ex 4.2 | Question 3

## 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.

Video Solution
Linear Equations In Two Variables
Ex 4.2 | Question 3

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