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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