**(1) 2x + 3y = 12; x – y = 1**

**Solution:**

Given

2x + 3y = 12

Substitute the values for x as shown in the table

x | 0 | 6 | 3 |

y | 4 | 0 | 2 |

Consider x – y = 1

x | 1 | 0 | 2 |

y | 0 | -1 | 1 |

Now plot the graph as shown below:

**(2) x – 3y = 1; 3x – 2y + 4 = 0**

**Solution:**

Given

x – 3y = 1

Substitute the values for x as shown in the table

x | -2 | 4 | 1 |

y | -1 | 1 | 0 |

Also given

3x – 2y + 4 = 0

x | 0 | -2 | -4 |

y | 2 | -1 | -1 |

Now plot the graph as shown below:

**(3) 5x – 6y + 30 = 0; 5x + 4y – 20 = 0**

**Solution:**

Given

5x – 6y + 30 = 0

Substitute the values for x as shown in the table

x | 0 | -6 | 6 |

y | 5 | 0 | 10 |

Also given

5x + 4y – 20 = 0

x | 0 | 4 | 8 |

y | 5 | 0 | -5 |

Now plot the graph as shown below:

**(4) 3x – y – 2 = 0; 2x + y = 8**

**Solution:**

For equation 1, let’s find the points for graph

3x – y – 2 = 0

At x = 0

3(0) – y – 2 = 0

⇒ y = -2

At x = 13

(1) – y – 2 = 0

⇒ y = 1At x = 2

3(2) – y – 2 = 0

⇒ 6 – y – 2 = 0⇒ y =4

Hence, points for graph are (0, -1) (1, 1) and (2, 4)

For equation 22x + y = 8

At x = 0y = 8

at x = 12

(1) + y = 8

⇒ y = 6

at x =42(4) + y = 8

⇒ y = 0

Hence, points for graph are (0, 8) (1, 6) and (4, 0)

From graph, we observe both lines intersect at (2, 4)

Hence, x = 2 y = 4 is the solution of given pair

**(5) 3x + y = 10; x – y = 2**

**Solution:**

Given 3x + y = 10

x | 1 | 2 | 3 |

y | 7 | 4 | 1 |

Also, we have

x – y = 2

x | 0 | 2 | 3 |

y | -2 | 0 | 1 |

Solving Both equations