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**Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 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)

x=142=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 a1a2,b1b2 and c1c2 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

a1a2=57b1b2=−46c1c2=8−9

Here, a1a2≠b1b2

∴ 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

a1a2=918=12b1b2=36=12

c1c2=1224=12

Here, a1a2=b1b2=c1c2

∴ 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

a1a2=62=31b1b2=−3−1=31

c1c2=109

Here, a1a2=b1b2≠c1c2

∴ Representation of lines graphically is parallel.

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

On compairing the ratios a1a2,b1b2 and c1c2 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) 32x+53y=7:9x−10y=14

iv) 5x – 3y = 11: -10x – 6y = -22

v) 43x+2y=8;2x+3y=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

a1a2=32b1b2=−23c1c2=−5−7=57

Here, a1a2≠b1b2

∴ 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

a1a2=24=12b1b2=−3−6=12

c1c2=−8−9=89

Here, a1a2=b1b2≠c1c2

∴ Graphical representation is parallel lines.

∴ Equations are inconsistent.

iii) 32x+53y=732x+53y−7=0

9x – 10y = 14 ⇒ 9x – 10y – 14 = 0

a1=32,b1=53,c1=−7

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

a1a2=32×19b1b2=53×−16

c1c2=−7−14=714=12

Here, a1a2≠b1b2

∴ 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

a1a2=5−10=−12b1b2=−36=−12

c1c2=−1122=−12

Here, a1a2=b1b2=c1c2

∴ Pair of equations are consistent

∴ Graphical representation is coninciding.

v) 43x+2y=843x+26−8=0

2x + 3y = 12 ⇒ 2x + 3y – 12 = 0

a1=43,b1=2,c1=−8

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

a1a2=43×12=16b1b2=23

c1c2=−8−12=23

Here, a1a2≠b1b2

∴ 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

a1a2=12b1b2=12c1c2=−5−10=12

Here, a1a2=b1b2=c1c2

∴ 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, a1a2=13b1b2=−1−3=13

c1c2=−8−16=12

a1a2=b1b2≠c1c2

∴ 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

y=−16+3×3

x | 6 | 8 |

y=−16+3×3 | 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

a1a2=24=12b1b2=1−2

c1c2=−6−4=32

Here, a1a2≠b1b2

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

y=−4+4×2

x | 1 | 3 |

y=−4+4×2 | 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

a1a2=24=12b1b2=−2−3=23

c1c2=−2−5=25

Here, a1a2≠b1b2

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

y=−2+2×2

∴ 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

y=−5+4×3

x | 2 | 5 |

y=−5+4×3 | 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.

2x+2y2=36

2x + 2y = 72

x + y = 36 …………… (ii)

∴ x – y = 4 (i)

x + y = 36 (ii)

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

x=402

∴ 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

a1a2=23b1b2=32

Here, when a1a2≠b1b2 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

a1a2=22=11b1b2=33=11

c1c2=−8−12=812=23

Here, a1a2=b1b2≠c1c2 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

a1a2=24=12b1b2=36=12

c1c2=−8−16=12

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

y=−3x+122

x | 0 | 2 |

y=−3x+122 | 6 | 3 |

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

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

**All Chapter KSEEB Solutions For Class 10 Maths**

—————————————————————————–**All Subject KSEEB Solutions For Class 10**

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