**KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2** 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 3.2.

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

**Pair Of Linear Equations In Two Variables Class 10 Notes Exercise 3.2 Question 1.**

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.

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

Solution:

(i) Strength of the X Std. is 10

Number of boys be ‘y’, then

number of girls be ‘x’ .

x + y = 10 …….. (i)

x = y + 4

∴ x – y = 4 ………. (ii)

From equations (i) + (ii)

\(\quad x=\frac{14}{2}=7\)

Substituting the value of ‘x’ in eqn. (i),

x + y = 10

7 + y = 10

y = 10 – 7

y = 4.

∴ Number of girls, x = 7

Number of boys, y = 4.

(ii) Cost of each pencil be Rs. ‘x’

Cost of pen be Rs. ‘y’

5x + 7y = 50 ………. (i)

7x + 5y = 46 ……….. (ii)

From equation (i) + equation (ii)

∴ x + y = 8 …………… (iii)

from Eqn. (ii) – Eqn. (i),

∴ -x + y = 2 …………. (iv)

Eqn. (iii) + Eqn. (iv)

∴ y = 5

Substituting the value of ‘y’ in eqn. (i)

x + y = 8

x + 5 = 8

∴ x = 8 – 5 x = 3

∴Cost of each pencil is Rs. 3.

Cost of each pen is Rs. 5.

**KSEEB Solutions For Class 10 Maths Pair Of Linear Equations In Two Variables Question 2.**

On comparing the ratios \(\frac{a_{1}}{a_{2}}, \frac{b_{1}}{b_{2}} \text { and } \frac{c_{1}}{c_{2}}\) 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

ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Solution:

i) 5x – 4y + 8 = 0 a_{1}= 5, b_{1} = -4, C_{1}= 8

7x + 6y – 9 = 0 a_{2} = 7, b_{2}= 6, c_{2}= -9

\(\frac{a_{1}}{a_{2}}=\frac{5}{7} \quad \frac{b_{1}}{b_{2}}=\frac{-4}{6} \quad \frac{c_{1}}{c_{2}}=\frac{8}{-9}\)

Here, \(\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}\)

∴ Lines representing the pairs of linear equations intersect at a point.

ii) 9x + 3y + 12 = 0

18x + 6y + 24 = 0

a_{1} =9, b_{1} = 3, c_{1} = 12

a_{2} = 18, b_{2} = 6, c_{2} = 24

\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{9}{18}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{6}=\frac{1}{2}\)

\(\frac{c_{1}}{c_{2}}=\frac{12}{24}=\frac{1}{2}\)

Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)

∴ Representation of lines graphically are coincident.

iii) 6x – 3y + 10 = 0

2x – y + 9 = 0

Here, a_{1} = 6, b_{1} = -3, c_{1} = 10

a_{2} = 2, b_{2} = -1, c_{2} = 9

\(\frac{a_{1}}{a_{2}}=\frac{6}{2}=\frac{3}{1} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-1}=\frac{3}{1}\)

\(\frac{c_{1}}{c_{2}}=\frac{10}{9}\)

Here, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}\)

∴ Representation of lines graphically is parallel.

**KSEEB Solutions For Class 10 Maths Question 3.**

On compairing the ratios \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}, \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}\) and \(\frac{c_{1}}{c_{2}}\) find out whether the following pair of linear equations are consistent, or inconsistent.

i) 3x + 2y = 5; 2x – 3y = 7

ii) 2x – 3y = 8; 4x-6y = 9

iii) \(\frac{3}{2} x+\frac{5}{3} y=7 : 9 x-10 y=14\)

iv) 5x – 3y = 11: -10x – 6y = -22

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

Solution:

i) 3x + 2y = 5 ⇒ 3x + 2y – 5 = 0

2x – 3y = 7 ⇒ 2x – 3y – 7 = 0

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

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

\(\frac{a_{1}}{a_{2}}=\frac{3}{2} \quad \frac{b_{1}}{b_{2}}=-\frac{2}{3} \quad \frac{c_{1}}{c_{2}}=\frac{-5}{-7}=\frac{5}{7}\)

Here, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

∴ Graphical representation is intersecting lines.

∴ Pair of linear equations are consistent.

ii) 2x – 3y = 8 ⇒ 2x – 3y – 8 = 0

4x – 6y = 9 ⇒ 4x – 6y – 9 = 0

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

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

\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{-6}=\frac{1}{2}\)

\(\frac{c_{1}}{c_{2}}=\frac{-8}{-9}=\frac{8}{9}\)

Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)

∴ Graphical representation is parallel lines.

∴ Equations are inconsistent.

iii) \(\frac{3}{2} x+\frac{5}{3} y=7 \quad \frac{3}{2} x+\frac{5}{3} y-7=0\)

9x – 10y = 14 ⇒ 9x – 10y – 14 = 0

\(a_{1}=\frac{3}{2}, \quad b_{1}=\frac{5}{3}, \quad c_{1}=-7\)

a_{2} = 9, b_{2} = -10, c_{2} = -17

\(\frac{a_{1}}{a_{2}}=\frac{3}{2} \times \frac{1}{9} \quad \frac{b_{1}}{b_{2}}=\frac{5}{3} \times \frac{-1}{6}\)

\(\frac{c_{1}}{c_{2}}=\frac{-7}{-14}=\frac{7}{14}=\frac{1}{2}\)

Here, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

∴ Pair of equations are consistent

iv) 5x – 3y = 11 ⇒ 5x – 3y – 11 = 0

-10x + 6y = – 22 ⇒ -10x + 6y + 22 = 0

a_{1} = 5, b_{1} = -3, c_{1} = -11

a_{2} = -10, b_{2} = 6, c_{2} = -22

\(\frac{a_{1}}{a_{2}}=\frac{5}{-10}=-\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{-3}{6}=-\frac{1}{2}\)

\(\frac{c_{1}}{c_{2}}=\frac{-11}{22}=-\frac{1}{2}\)

Here, \(\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)

∴ Pair of equations are consistent

∴ Graphical representation is coninciding.

v) \(\frac{4}{3} x+2 y=8 \quad \frac{4}{3} x+26-8=0\)

2x + 3y = 12 ⇒ 2x + 3y – 12 = 0

\(a_{1}=\frac{4}{3}, \quad b_{1}=2, \quad c_{1}=-8\)

a_{2} = 2, b_{2} = 3, c_{2} = -12

\(\frac{a_{1}}{a_{2}}=\frac{4}{3} \times \frac{1}{2}=\frac{1}{6} \quad \frac{b_{1}}{b_{2}}=\frac{2}{3}\)

\(\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{2}{3}\)

Here, \(\frac{\mathbf{a}_{1}}{\mathbf{a}_{2}} \neq \frac{\mathbf{b}_{1}}{\mathbf{b}_{2}}\)

∴ Pair of equations are consistent

**Pair Of Linear Equations Exercise 3.2 KSEEB Solutions Question 4.**

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 – 3y – 5 = 0

Solution:

(i) x + y = 5 ⇒ x + y – 5 = 0

2x + 2y = 10 ⇒ 2x + 2y – 10 = 0

a_{1} = 1, b_{1} = 1, c_{1} = -5

a_{2} = 2, b_{2} = 2, c_{2} = -10

\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{2} \quad \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}=\frac{-5}{-10}=\frac{1}{2}\)

Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)

∴ Pair of equations are consistent

(i) x + y = 5

y = 5 – x

x | 0 | 2 | 4 |

y = 5 – x | 5 | 3 | 1 |

(ii) 2x + 2y = 10

x + y = 5

y = 5 – x

x | 0 | 2 | 5 |

y = 5 – x | 5 | 3 | 0 |

∴ We can give any value for ‘x’, i.e., solutions are infinite.

∴ Solution, P (5, 0) x = 5, y = 0

(ii) x – y = 8 ⇒ x – y – 8 = 0

3x – 3y = 16 ⇒ 3x – 3y – 16 = 0

Here, \(\frac{a_{1}}{a_{2}}=\frac{1}{3} \quad \frac{b_{1}}{b_{2}}=\frac{-1}{-3}=\frac{1}{3}\)

\(\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}\)

\(\quad \frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\)

∴ Linear equations are inconsistenttent.

∴ Algebraically it has no solution.

Graphical representation → Parallel Lines.

(i) x – y = 8

-y = 8 – x

y = -8 + x

x | 8 | 10 | 9 |

y = -8+x | 0 | 2 | 1 |

(ii) 3x – 3y = 16

-3y = 16 – 3x

3y = -16 + 3x

\(\quad y=\frac{-16+3 x}{3}\)

x | 6 | 8 |

\(y=\frac{-16+3 x}{3}\) | 0.8 | 2.6 |

No solution because it is inconsistent

(iii) 2x + y – 6 = 0

4x – 2y – 4 = 0

Here a_{1} = 2, b_{1} = 1, c_{1} = -6

a_{2} = 4, b_{2} = -2, c_{2} = -4

\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{1}{-2}\)

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

Here, \(\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\)

Pair of equations are consistent. Algebraically both lines intersect.

Graphical Representation :

(i) 2x + y = 6

y = 6 – 2x

x | 0 | 2 |

y = 6 – 2x | 6 | 2 |

(ii) 4x – 2y – 4 = 0

4x – 2y = 4

-2y = 4 – 4x

2y = -4 + 4x

\(\quad y=\frac{-4+4 x}{2}\)

x | 1 | 3 |

\(y=\frac{-4+4 x}{2}\) | 0 | 4 |

Solution: intersecting point, P (2, 2) i.e., x = 2, y = 2

(iv) 2x – 2y – 2 = 0

4x – 3y – 5 = 0

a_{1} = 2, b_{1} = -2, c_{1} = -2

a_{2} = 4, b_{2} = -3, c_{2} = -5

\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{-2}{-3}=\frac{2}{3}\)

\(\frac{c_{1}}{c_{2}}=\frac{-2}{-5}=\frac{2}{5}\)

Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}\)

Pair of equations are consistent.

∴ Algebraically both lines intersect.

Graphical Representation :

(i) 2x – 2y – 2 =0

2x – 2y = 2

-2y = 2 – 2x

2y = -2 + 2x

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

∴ y = – 1 + x

x | 2 | 4 |

y= -1+ x | 1 | 3 |

(ii) 4x – 3y – 5 = 0

4x – 3y = 5

-3y = 5 – 4x

3y = -5 + 4x

\(\quad y=\frac{-5+4 x}{3}\)

x | 2 | 5 |

\(y=\frac{-5+4 x}{3}\) | 1 | 5 |

Solution: P(2, 1) i.e., x = 2, y = 1

**Pair Of Linear Equations In Two Variables KSEEB Solutions Question 5.**

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

Solution:

Length of rectangular garden be ‘x’ m.

Breadth of rectangular garden be ‘y’ m, then

Length of the garden is 4m more than its width.

x = y + 4 ……….. (i)

x – y = 4

Half of the circumference is 36 m.

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

2x + 2y = 72

x + y = 36 …………… (ii)

∴ x – y = 4 (i)

x + y = 36 (ii)

From eqn. (i) + eqn. (ii)

\(\quad x=\frac{40}{2}\)

∴ x = 20 m.

Substituting the value of ’x’ in eqn. (i)

x – y = 4

20 – y = 4

-y = 4 – 20

-y = -16

y = 16 m.

∴ Length of the garden = 20 m.

Breadth of the garden = 16 m.

**Pair Of Linear Equations Exercise 3.2 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 :

(i) intersecting lines

(ii) parallel lines

(iii) coincident lines.

Solution:

Linear equation is 2x + 3y – 8 = 0.

(i) Linear equations for intersecting lines:

2x + 3y – 8 = 0

3x + 2y – 7 = 0

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

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

\(\frac{a_{1}}{a_{2}}=\frac{2}{3} \quad \frac{b_{1}}{b_{2}}=\frac{3}{2}\)

Here, when \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}} \neq \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}\) Geometrical representation is intersecting lines.

(ii) For Parallel lines :

2x + 3y – 8 = 0

2x + 3y – 12 = 0

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

a_{1} = 2, b_{1} = 3, c_{1} = -12

\(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{2}{2}=\frac{1}{1} \quad \frac{\mathrm{b}_{1}}{\mathrm{b}_{2}}=\frac{3}{3}=\frac{1}{1}\)

\(\frac{c_{1}}{c_{2}}=\frac{-8}{-12}=\frac{8}{12}=\frac{2}{3}\)

Here, \(\frac{\mathrm{a}_{1}}{\mathrm{a}_{2}}=\frac{\mathrm{b}_{1}}{\mathrm{b}_{2}} \neq \frac{\mathrm{c}_{1}}{\mathrm{c}_{2}}\) hence

Graphical representation is parallel lines.

(iii) For Intersecting lines :

2x + 3y – 8 = 0

4x + 6y – 16 = 0

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

a_{1} = 4, b_{1} = 6, c_{1} = -16

\(\frac{a_{1}}{a_{2}}=\frac{2}{4}=\frac{1}{2} \quad \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}\)

\(\frac{c_{1}}{c_{2}}=\frac{-8}{-16}=\frac{1}{2}\)

Here,

∴ Lines are intersecting.

**10th Maths Pair Of Linear Equations In Two Variables Exercise 3.2 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 ………. (i)

3x – 2y – 12 = 0 ……….. (ii)

x – y + 1 = 0

-y = -x -1

y = x + 1

x | 2 | 4 |

y = x + 1 | 3 | 5 |

3x + 2y – 12 = 0

2y = -3x + 12

\(\quad y=\frac{-3 x+12}{2}\)

x | 0 | 2 |

\(y=\frac{-3 x+12}{2}\) | 6 | 3 |

Graphs of these two equations intersect at A. Vertices formed for ∆ABC are,

A (2, 3), B (-1, 0), C (4, 0).

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.2 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 3.2, drop a comment below and we will get back to you at the earliest.