# Class 10 Chapter 3 Exercise 3.1 Maths NCERT Solutions

Form the pair of linear equations in 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.

Let the no. of boys taking part be x

Let the no. of girls taking part be y

Since, total students taking part in quiz are 10.

So,

x + y = 10

y = 10 – x. —- (i)

A/Q,

y = x + 4. —–(ii)

Table for eq (i)

x | 4 | 5 | 6 |

y | 6 | 5 | 4 |

Table for eq (ii)

x | 0 | 1 | -1 |

y | 4 | 5 | 3 |

The point of intersection of both lines is (3,7)

So,

No. of boys taking part is x = 3 boys

No. of girls taking part is y = 7 girls

(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

let the cost of one pencil be ₹x

cost of one pen = ₹y

A/Q,

5x + 7y = 50

5x = 50 – 7y

x = $$\frac{50 -7y}{5}$$ —–(i)

and

7x + 5y =46

5y = 46 -7y

y = $$\frac{46 – 7x}{5}$$ —–(ii)

Table for eq (i)

x | 3 | 10 |

y | 5 | 0 |

Table for eq (ii)

x | 8 | 3 |

y | -2 | 5 |

The point of intersection of both lines is (3,5)

So,

Cost of one pencil = x = 3

Cost of one pen = y = 5

On comparing the ratios $$\frac{a_{1}}{a_{2}}$$,$$\frac{b_{1}}{b_{2}}$$,and $$\frac{c_{1}}{c_{2}}$$, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:

(i) 5x – 4y + 8 = 0

7x + 6y – 9 = 0

$$\frac{a_{1}}{a_{2}} = \frac{5}{7}$$

$$\frac{b_{1}}{b_{2}} = \frac{-4}{6}$$

$$\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$

So, the line are intersecting at a point

(ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

$$\frac{a_{1}}{a_{2}} = \frac{9}{18} = \frac{1}{2}$$

$$\frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}$$

$$\frac{c_{1}}{c_{2}} = \frac{12}{24} = \frac{1}{2}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$

So, the line are coincident

(iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

$$\frac{a_{1}}{a_{2}} = \frac{6}{2} = 3$$

$$\frac{b_{1}}{b_{2}} = \frac{-3}{-1} = 3$$

$$\frac{c_{1}}{c_{2}} = \frac{10}{9}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

So, the line are parallel

On comparing the ratios $$\frac{a_{1}}{a_{2}}$$,$$\frac{b_{1}}{b_{2}}$$,and $$\frac{c_{1}}{c_{2}}$$, find out whether the lines representing the following pair of linear equations are consistent, or inconsistent.

(i) 3x + 2y = 5 ; 2x – 3y = 7

$$\frac{a_{1}}{a_{2}} = \frac{3}{2}$$

$$\frac{b_{1}}{b_{2}} = \frac{2}{-3}$$

$$\frac{c_{1}}{c_{2}} = \frac{5}{7}$$

$$\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$

So, the line are intersecting at a point and has a solution. The lines are consistent.

(ii) 2x – 3y = 8 ; 4x – 6y = 9

$$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}$$

$$\frac{b_{1}}{b_{2}} = \frac{-3}{-6} = \frac{1}{2}$$

$$\frac{c_{1}}{c_{2}} = \frac{8}{9}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

So, the line are parallel and has no solution

$$\therefore$$ Lines are inconsistent

(iii) $$\frac{3}{2}x$$ + $$\frac{5}{3}y$$ = 7; 9x – 10y = 14

$$\frac{a_{1}}{a_{2}} = \frac{\frac{3}{2}}{9} = \frac{3}{2 \times 9} = \frac{1}{2\times3} = \frac{1}{6}$$

$$\frac{b_{1}}{b_{2}} = \frac{\frac{5}{3}}{-10} = \frac{5}{3 \times (-10)} = \frac{-1}{3 \times 2} = \frac{-1}{6}$$

$$\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$

So, the line are intersecting at a point and has solution

$$\therefore$$ Lines are consistent

(iv) 5x – 3y = 11 ; – 10x + 6y = –22

$$\frac{a_{1}}{a_{2}} = \frac{5}{-10} = \frac{-1}{2}$$

$$\frac{b_{1}}{b_{2}} = \frac{-3}{6} = \frac{-1}{2}$$

$$\frac{c_{1}}{c_{2}} = \frac{11}{-22} = \frac{-1}{2}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$

So, the line are coincident and has infinitely many solutions

$$\therefore$$ Lines are consistent

(v) $$\frac{4}{3}x$$ + 2y = 8; 2x + 3y = 12

$$\frac{a_{1}}{a_{2}} = \frac{\frac{4}{3}}{2} = \frac{4}{3 \times 2} = = \frac{2}{3}$$

$$\frac{b_{1}}{b_{2}} = \frac{2}{3}$$

$$\frac{c_{1}}{c_{2}} = \frac{8}{12} = \frac{2}{3}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$

So, the line are coincident

$$\therefore$$ Lines are consistent

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

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

$$\frac{a_{1}}{a_{2}} = \frac{1}{2}$$

$$\frac{b_{1}}{b_{2}} = \frac{1}{2}$$

$$\frac{c_{1}}{c_{2}} = \frac{5}{10} = \frac{1}{2}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$

So, the line are coincident and has infinitely many solutions

$$\therefore$$ Lines are consistent

y = 5 – x —-(i)

Table for eq(i)

x | 2 | 3 |

y | 3 | 2 |

y = $$\frac{10 – 2x}{2}$$ —-(ii)

Table for eq(ii)

x | 4 | 5 |

y | 1 | -0 |

(ii) x – y = 8, 3x – 3y = 16

$$\frac{a_{1}}{a_{2}} = \frac{1}{3} = \frac{1}{3}$$

$$\frac{b_{1}}{b_{2}} = \frac{-1}{-3} = \frac{1}{3}$$

$$\frac{c_{1}}{c_{2}} = \frac{8}{16} = \frac{1}{2}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

So, the line are parallel and has no solution

$$\therefore$$ Lines are inconsistent

(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0

$$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}$$

$$\frac{b_{1}}{b_{2}} = \frac{1}{-2} = \frac{-1}{2}$$

$$\frac{c_{1}}{c_{2}} = \frac{-6}{-4} = \frac{3}{2}$$

$$\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$

So, the line are intersecting and has solution

$$\therefore$$ Lines are consistent

y = 6 – 2x

Table for eq(i)

x | 2 | 1 |

y | 2 | 4 |

2y = 4x – 4

y = $$\frac{4x – 4}{2}$$

Table for eq(ii)

x | 1 | 0 |

y | 0 | -2 |

(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

$$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}$$

$$\frac{b_{1}}{b_{2}} = \frac{-2}{-4} = \frac{1}{2}$$

$$\frac{c_{1}}{c_{2}} = \frac{-2}{-5} = \frac{2}{5}$$

$$\because \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

So, the line are parallel and has no solution

$$\therefore$$ Lines are inconsistent

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.

let the length of garden be x m

the width of garden = y m

A/Q,

length is 4 more than its width

x = 4 + y

x – y = 4 ——(i)

and

$$\frac{2(x + y)}{2}$$ = 36

x + y = 36 ——(ii)

Adding eq (i) & (ii)

x – y = 4 |

x + y = 36 |

2x = 40 |

x = 20 m

Putting value of x in eq (i)

20 – y = 4

y = 16

So, length = x = 20 m

width = y= 16 m

Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

(i) intersecting lines

Intersecting lines :- for having intersecting lines

$$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$$

$$\therefore$$ Required linear equation is 5x – 6y + 2 = 0

(ii) parallel lines

Parallel lines :- for having parallel lines

$$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$$

$$\therefore$$ Required linear equation is 4x + 6y + 8 = 0

(iii) coincident lines

Coincident lines :- for having coincident lines

$$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$$

$$\therefore$$ Required linear equation is 8x + 12y – 32 = 0

Draw the graphs of the equations x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

x – y + 1 = 0

x = y – 1 —-(i)

Table for eq(i)

x | 0 | 1 |

y | 1 | 2 |

3x + 2y – 12 = 0

2y = 12 – 3x

y = $$\frac{12 – 3x}{2}$$ —-(i)

Table for eq(ii)

x | 4 | 2 |

y | 0 | 3 |

Coordinates of triangle formed are (-1,0), (2,3) and (4,0)