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

Go back to  'Ex.4.3'

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.