**KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1.

## Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

Question 1.

Aftab tells his daughter, “Seven years ago, I was, seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.

Solution:

Let the present age of father be x’

Let the age of daughter be ‘y’.

Before seven years, age of father is x – 7

Before seven years, age of daughter is y – 7.

Father was 7 times as old as daughter.

∴ x – 7 = 7 (y – 7)

x – 7 = 7y – 49

x – 7y = -49 + 7

x – 7y = – 42

∴ x – 7y + 42 = 0 …………….. (i)

After three years, age of father is x + 3

After three years, age of daughter is y + 3

Three years from now will three times as old as father.

∴ x + 3 = 3(y + 3)

x + 3 = 3y + 9

x – 3y + 3 – 9 = 0

3 – 3y – 6 = 0 ………….. (ii)

∴ Algebraically linear equations :

x – 7y + 42 = 0

x – 3y – 6 =0

If we represent this linear equations through graph, we have

i) x – 7y + 42 = 0

-7y = -x – 42

7y = x + 42

\(\quad y=\frac{x+42}{7}\)

x | 0 | 7 |

\(y=\frac{x+42}{7}\) | 6 | 7 |

ii) x – 3y – 6 = 0

-3y = -x + 6

3y = x – 6

\(\quad y=\frac{x-6}{3}\)

x | 0 | +6 |

\(y=\frac{x-6}{3}\) | -2 | 0 |

Question 2.

The coach of a cricket team buys 3 bats and 6 balls for Rs. 3900. Later, she buys another bat and 3 more balls of the same kind for Rs. 1300. Represent this situation algebriacally and geometrically.

Solution:

Let the cost of each bat be Rs. ‘x’

Let the cost of each ball be Rs. ‘y’.

3x + 6y = 3900

∴ x + 2y = 1300 ………. (i)

Cost of another bat and ball be x + 3y = 1300 ………….. (ii)

∴Algebriacally

x + 2y = 1300

x + 3y = 1300

And to represent through graph :

i) x + 2y = 1300

2y = 1300 – x

\(\quad x=\frac{1300-x}{2}\)

x | 100 | 200 |

\(y=\frac{1300-x}{2}\) | 600 | 500 |

ii) x + 3y = 1300

3y = 1300 – x

\(\quad y=\frac{1300-x}{3}\)

x | 100 | 400 |

\(y=\frac{1300-x}{3}\) | 400 | 900 |

Question 3.

The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs. 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs. 300. Represent the situation algebraically and geometrically.

Solution:

Algebraically,

If Cost of each kg. of apple be Rs. ‘x’,

Cost of grapes be Rs. ‘y’. Then

2x + y = 160

4x + 2y = 300

To represent geometrically,

i) 2x + y = 160

y = -2x + 160

x | 20 | 80 |

y = 2x + 160 | 120 | 0 |

ii) 4x + 2y = 300

2x + y= 150

y = 150 – 2x

x | 40 | 60 |

y = 150 – 2x | 70 | 30 |

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1, drop a comment below and we will get back to you at the earliest.