# Ex.4.2 Q2 Linear Equations in Two Variables Solution - NCERT Maths Class 9

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## Question

Write four solutions for each of the following equations:

(i) $$2 \mathrm{x}+\mathrm{y}=7$$

(ii) $$\pi x+y=9$$

(iii) $$x=4 y$$

Video Solution
Linear Equations In Two Variables
Ex 4.2 | Question 2

## Text 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}

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