**NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables**

NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables will give you an idea of solving two linear equations in two variables. In the questions of the exercise of chapter 3 -Linear Equation In two variables two conditions of two variables would be given and you are required to build two relations between the two variables in the form of two equations and then solving them for two or more than 2 values and draw the graphs wherever both graphs intersect each other,the corresponding values of x and y of the intersecting point is the solution of the linear equation.

**NCERT Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables**

**Exercise 3.1 -Linear Equations in Two Variables**

**Exercise 3.2 -Linear Equations in Two Variables**

**Class 10 maths NCERT solutions of important questions of chapter 3 Pair of Linear Equations**

**Exercise 3.7 – Linear Equations in Two Variables**

**NCERT Solutions for Class 10 Maths Exercise 3.1 of Chapter 3 Linear Equations in Two Variables**

**Q1.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 be interesting?)Represent this situation algebraically and graphically.**

Ans. Let the age of Aftab is x and of his daughter is y

According to first condition, seven years ago the age of Aftab = x – 7 and of his daughter =y-7,further it is given that

Seven years back the age of Afteb = 7× Seven years back his daughter’s age

x – 7 = 7(y-7)

x -7 = 7y -49

x – 7y = -49 + 7

x – 7y = -42……(i)

According to second condition,after 3 years the age of Aftab is = x + 3 and of his daughter’s age = y +3,furher it is given that

After 3 years the age of Aftab = 3× His daughter’s age after 3 years

x + 3 = 3(y + 3)

x + 3 = 3y + 9

x – 3y = 6……(ii)

Solutions of the equation x – 7y = -42

x | 0 | 7 | -7 |

y | 6 | 7 | 5 |

Solutions of the equation x – 3y = 6

x | 0 | 6 | -3 |

y | -2 | 0 | -3 |

Graphical representation of both equations

**Q2.The coach of a cricket team buys 3 bats and 6 balls for ₹ 3900. Later, she buys another bat and 3 more balls of the same kind for ₹1300. Represent this situation algebraically and geometrically.**

Ans. Let the cost of 1 bat is x and of 1 ball is y

According to fist condition

3 × cost of 1 bat + 6× cost of 1 ball = Rs 3900

3x + 6y = 3900…..(i)

According to second condition

Cost of 1 bat+ 3× cost of 1 ball = Rs 1300

x + 3y = 1300….(ii)

Solutions of equation (i) 3x + 6y = 3900

x | 100 | -100 | 300 |

y | 600 | 700 | 500 |

Solutions of equation (ii) x + 3y = 1300

x | 100 | 400 | -200 |

y | 400 | 300 | 500 |

Graphical representation of both equations

**Q3.The cost of 2 kg of apples and 1 kg of grapes on a day was found to be ₹160. After a month, the cost of 4 kg of apples and 2 kg of grapes is ₹300. Represent the situation algebraically and geometrically.**

Ans. Let the cost of 1 kg apple is = x and the cost of 1 kg grapes is= y

According to first condition of the question

2×cost of 1 kg apple +cost of 1 kg grapes = Rs 160

2x + y = 160…..(i)

According to second condition of the question

4×cost of 1 kg apple +2×cost of 1 kg grapes = Rs 300

**4x + 2y = 300**…..(ii)

Solutions of the first equation 2x + y = 160

x | 80 | 60 | 40 |

y | 0 | 40 | 80 |

Solutions of the second equation 4x + 2y = 300

x | 60 | 40 | 20 |

y | 30 | 70 | 110 |

Graphical representation of both equations

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**NCERT Solutions of Science and Maths for Class 9,10,11 and 12**

**NCERT Solutions of class 9 maths**

Chapter 1- Number System | Chapter 9-Areas of parallelogram and triangles |

Chapter 2-Polynomial | Chapter 10-Circles |

Chapter 3- Coordinate Geometry | Chapter 11-Construction |

Chapter 4- Linear equations in two variables | Chapter 12-Heron’s Formula |

Chapter 5- Introduction to Euclid’s Geometry | Chapter 13-Surface Areas and Volumes |

Chapter 6-Lines and Angles | Chapter 14-Statistics |

Chapter 7-Triangles | Chapter 15-Probability |

Chapter 8- Quadrilateral |

**NCERT Solutions of class 9 science **

**CBSE Class 9-Question paper of science 2020 with solutions**

**CBSE Class 9-Sample paper of science**

**CBSE Class 9-Unsolved question paper of science 2019**

**NCERT Solutions of class 10 maths**

**CBSE Class 10-Question paper of maths 2021 with solutions**

**CBSE Class 10-Half yearly question paper of maths 2020 with solutions**

**CBSE Class 10 -Question paper of maths 2020 with solutions**

**CBSE Class 10-Question paper of maths 2019 with solutions**

**NCERT solutions of class 10 science**

**Solutions of class 10 last years Science question papers**

**CBSE Class 10 – Question paper of science 2020 with solutions**

**CBSE class 10 -Latest sample paper of science**

**NCERT solutions of class 11 maths**

Chapter 1-Sets | Chapter 9-Sequences and Series |

Chapter 2- Relations and functions | Chapter 10- Straight Lines |

Chapter 3- Trigonometry | Chapter 11-Conic Sections |

Chapter 4-Principle of mathematical induction | Chapter 12-Introduction to three Dimensional Geometry |

Chapter 5-Complex numbers | Chapter 13- Limits and Derivatives |

Chapter 6- Linear Inequalities | Chapter 14-Mathematical Reasoning |

Chapter 7- Permutations and Combinations | Chapter 15- Statistics |

Chapter 8- Binomial Theorem | Chapter 16- Probability |

**CBSE Class 11-Question paper of maths 2015**

**CBSE Class 11 – Second unit test of maths 2021 with solutions**

**NCERT solutions of class 12 maths**

Chapter 1-Relations and Functions | Chapter 9-Differential Equations |

Chapter 2-Inverse Trigonometric Functions | Chapter 10-Vector Algebra |

Chapter 3-Matrices | Chapter 11 – Three Dimensional Geometry |

Chapter 4-Determinants | Chapter 12-Linear Programming |

Chapter 5- Continuity and Differentiability | Chapter 13-Probability |

Chapter 6- Application of Derivation | CBSE Class 12- Question paper of maths 2021 with solutions |

Chapter 7- Integrals | |

Chapter 8-Application of Integrals |

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