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

## 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.