## Chapter 4 Linear Equations in Two Variables Exercise 4.2

**Question 1: Which one of the following options is true, and why?**

y = 3x + 5 has

(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

Answer:

y = 3x + 5 has

(i) a unique solution, (ii) only two solutions, (iii) infinitely many solutions

Answer:

(iii) infinitely many solutions is the correct option as a linear equation in two variables has many solutions i.e. infinite solutions.

**Question 2: Write four solutions for each of the following equations:**

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

Answer:

(i) 2x + y = 7 (ii) πx + y = 9 (iii) x = 4y

Answer:

i) 2x + y = 7

x |
2 |
1 |
3.5 |
0 |

y |
3 |
5 |
0 |
7 |

ii) πx + y = 9

x |
1 |
0 |
-1 |
9/π |

y |
π - 9 |
9 |
9 + π |
0 |

iii) x = 4y

x |
0 |
4 |
-4 |
2 |

y |
0 |
1 |
-1 |
1/2 |

**Question 3: Check which of the following are solutions of the equation x - 2y = 4 and which are not:(i) (0, 2) (ii) (2, 0) (iii) (4, 0) (iv) (√2 , 4√2) (v) (1, 1)Answer:**

For i) (0, 2)

x - 2y = 4

= (0) - 2(2) = 4

= 0 - 4 = 4

= -4 ≠ 4

For ii) (2, 0)

x - 2y = 4

= (2) - 2(0) = 4

= 2 - 0 = 4

= 2 ≠ 4

For iii) (4, 0)

x - 2y = 4

= (4) - 2(0) = 4

= 4 - 0 = 4

= 4 = 4

For iv) (√2 , 4√2)

x - 2y = 4

= (√2) - 2(4√2) = 4

= √2 - 8√2 = 4

= -7√2 ≠ 4

For v) (1, 1)

x - 2y = 4

= (1) - 2(1) = 4

= 1 - 2 = 4

= -1 ≠ 4

Therefore option iii) is solution for x - 2y = 4.

**Question 4: Find the value of k, if x = 2, y = 1 is a solution of the equation 2x + 3y = k.**

Answer:

Answer:

If x = 2 and y = 1, then the value of k in 2x + 3y = k is

2(2) + 3(1) = k

= 4 + 3 = k

= 7 = k

Hence k = 7.

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