Ex.4.3 Q1 Linear Equations in Two Variables Solution - NCERT Maths Class 9


Question

Draw the graph of each of the following linear equations in two variables:

(i) \(x + y = 4\)

(ii) \(x – y = 2\)

(iii) \(y = 3x\)

(iv) \(3 = 2x + y\)

 Video Solution
Linear Equations In Two Variables
Ex 4.3 | Question 1

Text Solution

What is known?

Linear equations

 What is Unknown?

How to draw graph

 Reasoning:

First of all, we can draw table for different values of \(x\) and \(y\) and then with the help of the values we can plot graph for each equation.

Steps:

(i) \(x + y = 4\)

Re-write the equation as

\(\begin{align}&x + y = 4 \\&\Rightarrow y = 4 \,– x \quad\dots\text{Equation } (1)\end{align}\)

By substituting the different values of \(x\) in the Equation (1), we get different values for \(y\)

  • When \(x = 0,\) we have: \(y = 4 – 0 = 4\) 
  • When \(x = 2,\) we have: \(y = 4 – 2 = 2\) 
  • When \(x = 4,\) we have: \(y = 4 – 4 = 0\)

Thus, we have the following table with all the obtained solutions:

\(x\) \(0\) \(2\) \(4\)
\(y\) \(4\) \(2\) \(0 \)

By Plotting the points \((0, 4) (2, 2)\) and \((4, 0)\) on the graph paper and drawing a line joining the corresponding points, we obtain the Graph.

The graph of the line represented by the given equation is as shown.

(ii) \(x – y = 2\)

Re-write the equation as
\(\begin{align}&x - y = 2\\&\Rightarrow y = x \,– 2\; \quad  \dots\text{Equation } (1)\end{align}\)

By substituting the different values of \(x\) in the Equation (1) we get different values for \(y\)

  • When \(x = 0,\) we have \(y = 0\, – 2 =\, – 2\) 
  • When \(x = 2,\) we have \(y = 2 \,– 2 = 0 \)
  • When \(x = 4,\) we have \(y = 4 \,– 2 = 2\)

Thus, we have the following table with all the obtained solutions:

\(x\) \(0\) \(2\) \(4\)
\(y\) \(-2\) \(0\) \(2\)

By Plotting the points \((0, –2)\), (2, 0) and (4, 2) on the graph paper and drawing a line joining the corresponding points, we obtain the Graph.

The graph of the line represented by the given equation is as shown

(iii) \(y = 3x\)

By substituting the different values of \(x\) in the Equation (1) we get different values for \(y\)

  • When \(x = 0,\) we have: \(y = 3(0) = 0\)
  • When \(x = 1,\) we have: \(y = 3 (1) = 3\)
  • When \(x =\,–1,\) we have:

 \(y = 3(–1) = \,– 3\)

Thus, we have the following table with all the obtained solutions:

\(x\) \(0\) \(1\) \(-1\)
\(y\) \(0\) \(3\) \(-3\)

By Plotting the points \((0, 0), (1, 3)\) and

\((–1, –3)\) on the graph paper and drawing a line joining the corresponding points, we obtain the Graph.

The graph of the line represented by the given equation is as shown:

(iv) \(3 = 2x + y\)

 Re-write the equation as
\(\begin{align} y = 3 – 2x\; \quad  \dots\text{Equation } (1)\end{align}\)

By substituting the different values of \(x\) in the Equation (1) we get different values for \(y\)

  • When \(x = 0,\) we have:

\(y = 3 \,– 2(0) = 3 \,– 0 = 3 \)

  • When \(x = 3,\) we have:

\(y = 3 \,– 2(3) = 3 \,– 6 = \,– 3 \)

  • When \(x = \,– 1,\) we have:

\(y = 3 \,– 2 (–1) = 3 + 2 = 5\)

Thus, we have the following table with all the obtained solutions:

\(x\) 0 3 \(-1\)
\(y\) 3 \(-3\) 5

By Plotting the points \((0, 3)\), \((3, –3)\) and \((–1, 5)\) on the graph paper and drawing a line joining the corresponding points, we obtain the Graph.

The graph of the line represented by the given equation is as shown.

 Video Solution
Linear Equations In Two Variables
Ex 4.3 | Question 1
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