**NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables**

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**NCERT Solutions for Class 10 Maths Exercise 3.2 of Chapter 3 Linear Equations in Two Variables**

**Q1. Form the pair of linear equations of the following problems and find their solutions graphically:**

**(i) 10 students of class X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.**

Ans. Let the number of girls is x and the number of boys is y who participated in the quiz.

It is given that total number of the students who took part in the quiz =10

x + y = 10……(i)

According to the second condition of question

Numer of girls = 4 more than the number of boys =4 + Number of the boys

x = y + 4

Rearranging the equation

x – y =4……(ii)

Solutions of the first equation x + y = 10

x | 4 | 5 | 6 |

y | 6 | 5 | 4 |

Solutions of the second equation x – y =4

x | 0 | 4 | 1 |

y | -4 | 0 | -3 |

It is clear from the graph that both lines intersect at (7,3) there fore x =7 and y = 3

Therefore number of girls 7 and number of boys 3 participated in the quiz.

**(ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and that of one pen.**

Ans. Let the cost of one pen is x and of one pencil is y

According to first condition

The cost of one pencil × 5 + The cost of one pen ×7 = Rs 50

5x + 7y =50…..(i)

According to the second condition

The cost of one pencil × 7 + The cost of one pen ×5 = Rs 50

7x + 5y =46…..(ii)

Solutions of the first equation 5x + 7y =50

x | 3 | -4 | 10 |

y | 5 | 10 | 0 |

Solutions of the second equation 7x + 5y =46

x | 3 | -2 | 8 |

y | 5 | 12 | -2 |

Both of the lines intersects at (3,5), x =3 and y =5,therefore cost of one pencil is Rs 3 and cost of one pen is Rs 5.

**Q2. On comparing the ratios a _{1}/a_{2}, b_{1}/b_{2 }and c_{1}/c_{2}, find out whether the lines representing the following pairs of linear equations intersects at a point,are parallel or coincident:**

**(i) 5x – 4y + 8 = 0,7x + 6y -9 = 0**

**(ii)9x + 3y +12 = 0,18x + 6y +24 = 0**

**(iii)6x -3y +10 = 0, 2x -y +9 = 0**

Ans. The given pair of linear equation is 5x – 4y + 8 = 0,7x + 6y -9 = 0

Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0

a_{1} =5, b_{1}=-4, c_{1}= 8, a_{2}=7, b_{2}=6, c_{2}= -9

a_{1}/a_{2},=5/7, b_{1}/b_{2 }= -4/6 =-2/3,c_{1}/c_{2 }= 8/-9 =-8/9

Here a_{1}/a_{2}≠b_{1}/b_{2}

Therefore both lines intersects at a point.

(ii) The given pair of linear equation is 9x + 3y +12 = 0,18x + 6y +24 = 0

Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0

a_{1} =9, b_{1}=3, c_{1}= 12, a_{2}=18, b_{2}=6, c_{2}= 24

a_{1}/a_{2},=9/18= 1/2, b_{1}/b_{2 }=3/6 = 1/2,c_{1}/c_{2 }= 12/24 =1/2

Therefore the given pair of linear equations are coincident

(iii) The given pair of linear equation is 6x -3y +10 = 0, 2x -y +9 = 0

Comparing the given pair of linear equations with standard pair of linear equations

Therefore the given linear equations are parallel to each other

**Q3. On comparing the ratios a _{1}/a_{2}, b_{1}/b_{2 }and c_{1}/c_{2}, find out whether the following pairs of linear equations consistent or inconsistent .**

**(i) 3x + 2y =5, 2x -3y = 7**

**(ii)2x -3y = 8, 4x – 6y = 9**

**(iii)3x/2 +5y/3 = 7, 9x -10 y =14**

Ans.

(i) Rearranging the given pair of linear equation is 3x + 2y -5=0, 2x -3y – 7=0

Comparing the given pair of linear equations with standard pair of linear equations a_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0

a_{1} =3, b_{1}=2, c_{1}= -5, a_{2}=2, b_{2}=-3, c_{2}= -7

a_{1}/a_{2}=3/2, b_{1}/b_{2 }= 2/-3=-2/3,c_{1}/c_{2 }= -5/-7 =5/7

a_{1}/a_{2}≠b_{1}/b_{2 }

Therefore the given pair of linear equations has unique solutions,thus it is consistent.

(ii) Rearranging the given pair of linear equation is 2x -3y – 8=0, 4x – 6y – 9=0

_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0

a_{1} =2, b_{1}=-3, c_{1}= -8, a_{2}=4, b_{2}=-6, c_{2}= -9

a_{1}/a_{2}=2/4=1/2, b_{1}/b_{2 }= -3/-6=1/2,c_{1}/c_{2 }= -8/-9 =8/9

a_{1}/a_{2}=b_{1}/b_{2}≠c_{1}/c_{2}

Therefore the given pair of linear equations are parallel ,it has no solutions thus it is Inconsistent.

(iii) Rearranging the given pair of linear equation 3x/2 +5y/3-7=0, 9x -10 y -14=0

_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0

a_{1} =3/2, b_{1}=5/3, c_{1}= -7, a_{2}=9, b_{2}=-10, c_{2}= -14

a_{1}/a_{2}=3/(2×9) = 1/6, b_{1}/b_{2 }= 5/(3×-10)=5/-30 =-1/6,c_{1}/c_{2 }= -7/-14=1/2

a_{1}/a_{2}≠b_{1}/b_{2}

Therefore the given pair of linear equations has unique solutions, it is consistent.

**Q4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically.**

**(i) x + y = 5, 2x + 2y = 10**

**(ii) x – y = 8, 3x – 3y = 16**

**(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0**

**(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0**

Ans.(i) Arranging the given pair of the linear equation x + y – 5=0, 2x + 2y – 10 = 0

_{1}x+b_{1}y_{ }+ c_{1}=0 and a_{2}x+b_{2}y_{ }+ c_{2}=0

a_{1}/a_{2 }= 1/2, b_{1}/b_{2 }= 1/2, c_{1}/c_{2 }= -5/-10 = 1/2,

∴ a_{1}/a_{2 }= b_{1}/b_{2 }= c_{1}/c_{2}

Therefore the given linear equations are coincident,so for every value of x there are infinite values of y

Solutions of the equation (i) x + y – 5=0

x | 1 | 4 | 2 | 3 |

y | 4 | 1 | 3 | 2 |

Solutions of the equation (ii) 2x + 2y – 10 = 0

x | 1 | 4 | 2 | 3 |

y | 4 | 1 | 3 | 2 |

Similarly (ii),(iii),(iv) and v part of the questions can be solved

**Q5.Half the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden graphically.**

Ans.According to first condition of the question,the half of the perimeter of rectangle = 36

Let the length of the rectangle = x and the length of the breadh =y

Half of the perimeter = 36

2(x +y)/2 = 36

x + y = 36……..(i)

According to second condition,the length is 4 m more than the breadth

Length = Width + 4 m

x = y + 4

arranging the equation

x – y = 4…….(ii)

Solutions of the first equation x + y = 36

x | 20 | 16 | 12 |

y | 16 | 20 | 24 |

Solutions of the second equation x – y = 4

x | 4 | 0 | 8 |

y | 0 | -4 | 4 |

Graphical representation of both equations

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