Jharkhand Board JAC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 Textbook Exercise Questions and Answers.

## JAC Board Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.2

Question 1.

Form the pair of linear equations in the following problems, and find their solutions graphically:

1. 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.

2. 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.

Solution:

1. Let the number of boys be x and the number of girls be y.

Then, the equations formed as follows:

x + y = 10 ……. (1)

and y = x + 4,

i.e., y – x = 4 …… (2)

To draw the graphs of these equations,

we find two solutions for each equation.

For equation (1), x + y = 10 gives y = 10 – x.

x | 0 | 5 |

y | 10 | 5 |

For equation (2), y – x = 4 gives y = x + 4.

x | 0 | 2 |

y | 4 | 6 |

Two lines intersect at point (3, 7). Hence, x = 3 and y = 7 is the required solution of the pair of linear equations.

Thus, 3 boys and 7 girls took part in the quiz.

Verification : x = 3 and y = 7 satisfy both the equations x + y = 10 and y – x = 4.

2. Let the cost of each pencil be ₹ x and the cost of each pen be ₹ y.

Then, from the given information, we receive the following equations:

5x + 7y = 50 ……… (1)

7x + 5y = 46 ……….. (2)

To draw the graphs of these equations, we find two solutions for each equation.

For equation (1), 5x + 7y = 50 gives

y = \(\frac{50-5 x}{7}\)

x | 3 | 10 |

y | 5 | 0 |

For equation (2), 7x + 5y = 46 gives

y = \(\frac{46-7 x}{5}\)

x | 3 | 8 |

y | 5 | -2 |

Two lines intersect at point (3, 5). Hence, x = 3 and y = 5 is the required solution of the pair of linear equations.

Thus, the cost of each pencil is ₹ 3 and the cost of each pen is ₹ 5.

Verification : x = 3 and y = 5 satisfy both the equations 5x + 7y = 50 and 7x + 5y = 46.

Question 2.

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:

1. 5x – 4y + 8 = 0; 7x + 6y – 9 = 0

2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

3. 6x – 3y + 10 = 0; 2x – y + 9 = 0

Solution:

5x – 4y + 8 = 0; 7x + 6y – 9 = 0

For the given pair of linear equations,

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

Now, \(\frac{a_1}{a_2}=\frac{5}{7}, \quad \frac{b_1}{b_2}=\frac{-4}{6}=-\frac{2}{3}\)

and \(\frac{c_1}{c_2}=\frac{8}{-9}=-\frac{8}{9}\)

Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

Hence, the lines representing the given pair of linear equations intersect at a point.

2. 9x + 3y + 12 = 0; 18x + 6y + 24 = 0

For the given pair of linear equations, a_{1} = 9, b_{1} = 3, c_{1} = 12, a_{2} = 18, b_{2} = 16 and c_{2} = 24.

Now, \(\frac{a_1}{a_2}=\frac{9}{18}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{3}{6}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{12}{24}=\frac{1}{2}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Hence, the lines representing the given pair of linear equations are coincident lines.

3. 6x – 3y + 10 = 0; 2x – y + 9 = 0

For the given pair of linear equations, a_{1} = 6, b_{1} = -3, c_{1} = 10, a_{2} = 2, b_{2} = -1 and c_{2} = 9.

Now, \(\frac{a_1}{a_2}=\frac{6}{2}=3, \quad \frac{b_1}{b_2}=\frac{-3}{-1}=3\)

and \(\frac{c_1}{c_2}=\frac{10}{9}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Hence, the lines representing the given pair of linear equations are parallel lines.

Question 3.

On comparing the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}\) and \(\frac{c_1}{c_2}\), find out whether the following pair of linear equations are consistent or inconsistent:

1. 3x + 2y = 5; 2x – 3y = 7

2. 2x – 3y = 8; 4x – 6y = 9

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

4. 5x – 3y = 11; -10x + 6y = -22

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

Solution:

1. 3x + 2y = 5; 2x – 3y = 7

For the given pair of linear equations, a_{1} = 3, b_{1} = 2, c_{1} = -5, a_{2} = 2, b_{2} = -3 and c_{2} = -7.

Now, \(\frac{a_1}{a_2}=\frac{3}{2}, \frac{b_1}{b_2}=\frac{2}{-3}=-\frac{2}{3}\)

and \(\frac{c_1}{c_2}=\frac{-5}{-7}=\frac{5}{7}\)

Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

Hence, the given pair of linear equations is consistent.

2. 2x – 3y = 8: 4x – 6y = 9

For the given pair of linear equations,

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

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{-6}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-8}{-9}=\frac{8}{9}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Hence, the given pair of linear equations is inconsistent.

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

Multiplying the first equation by 6 and expressing both the equations in the standard form, we get following equations:

9x + 10y – 42 = 0; 9x – 10y – 14 = 0

For the given pair of linear equations, a_{1} = 9, b_{1} = 10, c_{1} = -42, a_{2} = 9, b_{2} = -10 and c_{2} = -14.

Now, \(\frac{a_1}{a_2}=\frac{9}{9}=1, \frac{b_1}{b_2}=\frac{10}{-10}=-1\)

and \(\frac{c_1}{c_2}=\frac{-42}{-14}=3\)

Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

Hence, the given pair of linear equations is consistent.

4. 5x – 3y = 11; -10x + 6y = -22

For the given pair of linear equations, a_{1} = 5, b_{1} =-3, c_{1} = -11, a_{2} = -10, b_{2} = 6 and c_{2} = 22.

Now, \(\frac{a_1}{a_2}=\frac{5}{-10}=-\frac{1}{2}, \frac{b_1}{b_2}=\frac{-3}{6}=-\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-11}{22}=-\frac{1}{2}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Hence, the given pair of linear equations is consistent and dependent.

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

For the given pair of linear equations, a_{1} = \(\frac{4}{3}\), b_{1} = 2, c_{1} = -8, a_{2} = 2, b_{2} = 3 and c_{2} = -12.

Now, \(\frac{a_1}{a_2}=\frac{\frac{4}{3}}{2}=\frac{2}{3}, \quad \frac{b_1}{b_2}=\frac{2}{3}\)

and \(\frac{c_1}{c_2}=\frac{-8}{-12}=\frac{2}{3}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Hence, the given pair of linear equations is consistent and dependent.

Question 4.

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

1. x + y = 5; 2x + 2y = 10

2. x – y = 8; 3x – 3y = 16

3. 2x + y – 6 = 0; 4x – 2y – 4 = 0

4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0

Solution:

1. x + y = 5; 2x + 2y = 10

For the given pair of linear equations,

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

Now, \(\frac{a_1}{a_2}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-5}{-10}=\frac{1}{2}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Hence, the given pair of linear equations is consistent and dependent.

Now, we draw the graphs of both the equations.

x + y = 5 gives y = 5 – x.

x | 0 | 5 |

y | 5 | 0 |

2x + 2y = 10 gives y = \(\frac{10-2 x}{2}\)

x | 1 | 3 |

y | 4 | 2 |

Here, lines representing both the equations. are coincident. Hence, any point on the line gives a solution. In general, y = 5 – x, where x is any real number is a solution of the given pair of linear equations.

2. x – y = 8; 3x – 3y = 16

For the given pair of linear equations, a_{1} = 1, b_{1} = -1, c_{1} = -8, a_{2} = 3, b_{2} = -3 and c_{2} = -16.

Now, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{-1}{-3}=\frac{1}{3}\)

and \(\frac{c_1}{c_2}=\frac{-8}{-16}=\frac{1}{2}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Hence, the given pair of linear equations is inconsistent.

3. 2x + y – 6 = 0; 4x – 2y – 4 = 0

For the given pair of linear equations, a_{1} = 2, b_{1} = 1, c_{1} = -6, a_{2} = 4, b_{2} = -2 and c_{2} = -4.

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{1}{-2}=-\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-6}{-4}=\frac{3}{2}\)

Here, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

Hence, the given pair of linear equations is consistent.

Now, we draw the graphs of both the equations.

2x + y – 6 = 0 gives y = 6 – 2x.

x | 0 | 3 |

y | 6 | 0 |

4x – 2y – 4 = 0 gives y = \(\frac{4 x-4}{2}\) = 2x – 2.

x | 0 | 1 |

y | -2 | 0 |

Here, the lines intersect at point (2, 2). Hence, x = 2 and y = 2 is the unique solution of the given pair of linear equations.

4. 2x – 2y – 2 = 0; 4x – 4y – 5 = 0

For the given pair of linear equations, a_{1} = 2, b_{1} = -2, c_{1} = -2, a_{2} = 4, b_{2} = -4 and c_{2} = -5.

Now, \(\frac{a_1}{a_2}=\frac{2}{4}=\frac{1}{2}, \quad \frac{b_1}{b_2}=\frac{-2}{-4}=\frac{1}{2}\)

and \(\frac{c_1}{c_2}=\frac{-2}{-5}=\frac{2}{5}\)

Here, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\)

Hence, the given pair of linear equations is inconsistent.

Question 5.

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.

Solution:

Let the length and breadth of the rectangular garden be x m and y m respectively.

Then, from the given data, x = y + 4 and 36 = \(\frac{1}{2}\)[2(x + y)] i.e., x + y = 36 as the perimeter of a rectangle = 2 (length + breadth).

To draw the graphs, we find two solutions of each equation.

x = y + 4 gives y = x – 4

x | 8 | 24 |

y | 4 | 20 |

x + y = 36 gives y = 36 – x

x | 12 | 24 |

y | 24 | 12 |

Here, the lines intersect at point (20, 16). Hence, x = 20 and y = 16 is the unique solution of the pair of linear equations.

Thus, for the rectangular garden, length = 20 m and breadth = 16 m.

Question 6.

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:

1. intersecting lines

2. parallel lines

3. coincident lines

Solution:

1. For intersecting lines, \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 3x + 4y – 24 = 0. Here, \(\frac{a_1}{a_2}=\frac{2}{3}\) and \(\frac{b_1}{b_2}=\frac{3}{4}\) satisfying \(\frac{a_1}{a_2} \neq \frac{b_1}{b_2}\)

2. For parallel lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0. We can give another equation as 6x + 9y – 10 = 0.

Here, \(\frac{a_1}{a_2}=\frac{1}{3}, \quad \frac{b_1}{b_2}=\frac{1}{3}\) and \(\frac{c_1}{c_2}=\frac{4}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2} \neq \frac{c_1}{c_2}\).

3. For coincident lines, \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\). The given equation is 2x + 3y – 8 = 0.

We can give another equation as 10x + 15y – 40 = 0. Here, \(\frac{a_1}{a_2}=\frac{1}{5}\), \(\frac{b_1}{b_2}=\frac{1}{5}\) and \(\frac{c_1}{c_2}=\frac{1}{5}\) satisfying \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\).

Question 7.

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.

Solution:

x – y + 1 = 0 gives y = x + 1

x | -1 | 2 |

y | 0 | 3 |

3x + 2y – 12 = 0 gives y = \(\frac{12-3 x}{2}\)

x | 0 | 4 |

y | 6 | 0 |

The vertices of the triangle formed by the given lines and the x-axis are (-1, 0), (4, 0) and (2, 3).