Telangana SCERT Class 10 Maths Solution Chapter 4 Pair of linear equations in two variables Exercise 4.1

Telangana SCERT Solution Class X (10) Maths Chapter 4 Pair of linear equations in two variables Exercise 4.1

 

Pair of linear equations in two variables

Exercise – 4.1

(Q1) 

By comparing the nations a1/a2, b1/b2, c1/c2 state

Whether the liens represented by the following pairs of linear equations intersect at a point/are parallel/are coincident.

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

18x + 6y + 24 = 0

=> Solution: Given linear equations are,

9x + 3y + 12 = 0 = —— (1) and

18x + 6y + 24 = 0 ——– (2)

By comparing a1x + b1y + c1 = 0 and

A2x + b2y + c2 = 0

∴ a1 = 9, b1 = 3, c1 = 12

A2 = 18, b2 = 6, c2 = 24

∴ a1/a2 = 9/18 = 1/2, b1/b2 = 3/6 = 1/2, c1/c2 = 12/24 = 1/2

∴ a1/a2 = b1/b2 = c1/c2

If a1/a2 = b1/b2 = c1/c2 then two lines are

Coincident.

(c) 6x – 3y + 10 = 0

2x – y + 9 = 0

=> Solution: Given linear equations are,

6x – 3y + 10 = 0 ——— (1)

2x – y + 9 = 0 ——- (2)

By comparing equation (1) and (2), with

A1x + b1y + c1 = 0 and a2x + b2y + c2 = 0

∴ a1 = 6, b1 = -3, c1 = 10

A2 = 2, b2 = -1, c2 = 9

∴ a1/a2 = 6/2 = 3, b1/b2 = -3/-1 =3, c1/c2 = 10/9

∴ a1/a2 = b1/b2 ≠ c1/c2

If a1/a2 = b1/b2 ≠ c1/c2 then two lines are parallel to each other.

(Q2) Check whether the following equation are consistent or inconsistent, solve them graphically.

(i) 3x + 2y = 5, 2x – 3y = 7

=> Solution:

Given pair of equation are,

3x + 2y = 5 and 2x – 3y = 7 —— (2)

Comparing with a1x + b1y + c1 = 0 and

A2x + b2y + c2 = 0

∴ a1 = 3, b1 = 2, c1= 5

A2 = 2, b2 = -3, c2 = 7

∴ a1/a2 = 3/2, b1/b2 = 2/-3, c1/c2 = 5/7

∴ a1/a2 ≠ b1b2

∴ Given linear equations or lines are intersecting and consistent.

Multiply by equation (1) X 2, we get

2 X 3x + 2 X 2y = 2 X 5

6x + 4y = 10 —— (3)

Multiply by equation (2) X 3, we get

3 X 2x – 3 X 3y = 3 X 7

6x – 9y = 21 —— (4)

Subtracting eqn (3) from (4), we get,

6x = 174/13

X = 174/13X6

X = 29/13

4 (0) – 6y = 9

-6y = 9 – 0

-6y = 9

Y = 9/-6

Y = -3/2

Y = -1.5

∴ The lines are parallel and has no solution.

(iv) 5x – 3y = 11

-10x + 6y = -22

=> Solution: Given pairs of equations,

5x + 3y = 11

-10x + 6y = -22

Comparing with a1x + b1y = c1 and

A2x + b2y = c2

∴ a1 = 5   b1 = -3   c1 = 11

A2 = -10   b2 = 6    c2 = -22

∴ a1/a2 = 5/-10 = -1/2, b1/b2 = -3/6 = -1/2, c1/c2 = 11/-22 = 1/-2

∴ a1/a2 = b1/b2 = c1/c2

∴ The given linear equations or lines consistent.

∴ The given lines are coincident lines

∴ They have infinitely many solution

(v) 4/3 x + 2y = 8

2x + 3y = 12

=> Solution: Given pairs of equations are,

4/3 x + 2y = 8

4x + 6y/3 = 8

4x + 6y = 24 = ——- (1)

2x + 3y = 12 ——- (2)

Comparing with a1x + b1y = c1 and

A2x + b2y = c2

∴ a1 = 4   b1 = 6   c1 = 24

A2 = 2    b2 = 3    c2 = 12

∴ a1/a2 = 4/2 = 2   b1/b2 = 6/3 = 2   c1/c2 = 24/12 = 2

∴ a1/a2 = b1/b2 = c1/c2

∴ The given linear equ or liens are consistent

∴ The given lines are coincident lines

∴ They have infinitely many solutions

(vi) x + Y = 5

2x + 2y = 10

=> Solution: Given pairs of equations are,

x + y = 5

2x + 2y = 10

Comparing with a1x + b1y = c1

A2x + b2y = c2

∴ a1 = 1   b1 = 1   c1 = 5

A2 = 2    b2 = 2    c2 = 10

∴ a1/a2 = 1/2   b1/b2 = 1/2   c1/c2 = 5/10 = 1/2

∴ a1/a2 = b1/b2 = c1/c2

∴ The given lines are consistent.

∴ The given lines are coincident lines

∴ They have infinitely many solutions

(vii) x – y =8

=> Solution: Given pairs of equations are,

x – y = 8

3x – 3y = 16

Comparing with a1x + b1y = c1 and

A2x + b2y = c2

∴ a1 = 1   b1 = -1    c1 = 8

A2 = 3   b2 = -3    c2 = 16

∴ a1/a2 = 1/3    b1/b2 = 1/3     c1/c2 = 8/16 = 1/2

∴ a1/a2 = b1/b2 ≠ c1/c2

∴ given linear equations or lines are parallel and inconsistent.

(viii) 2x + y – 6 = 0

4x – 2y – 4 = 0

=> Solution: Given pairs of equations are,

2x + y = 6 —— (1)

4x – 2y = 4 —– (2)

Comparing with a1x + b1y = c1 and

A2x + b2y = c2

∴ a1 = 2    b1 = 1    c1 = 6

A2 = 4    b2 = -2    c2 = 4

∴ a1/a2 = 2/4 = 1/2    b1/b2 = -1/2   c1/c2 = 6/4 = 3/2

∴ a1/a2 ≠ b1/b2

∴ The given lines are consistent

∴ The lines interact at a point and gives only one solution.

 

(ix) 2x – 2y = 2

4x – 4y = 5

=> Solution: Given pairs of equations are,

2x – 2y = 2

4x – 4y = 5

Comparing with a1x + b1y = c1 and

A2x + b2y = c2

∴ a1 = 2    b1 = -2   c1 = 2

A2 = 4    b2 = -4    c2 = 5

∴ a1/a2 = 2/4 = 1/2    b1/b2 = -2/-4 = 1/2    c1/c2 = 2/5

∴ a1/a2 = b1/b2 ≠ c1/c2

∴ The given linear equor lines are parallel and inconsistent

∴ The given lines are parallel and has no solution.

(Q3) Form equations for the following situations and them graphically.

(1) Neha went to a shop to purchase some pants and skirts when her friends asked her how many of each she had bought, she answered, “the number of skirts are two less the twice the the number of pants purchased and the number of skirts is four less than four times the number of pants purchased”.

Help her friend to find how many pants and skirts Neha bought.

=> Solution:

Let x be the number of skirts and y be the number of pants.

From first condition,

X = 2y – 2

From second condition

X = 4y – 2

(ii) 10 students of class – x took part in a mathematics quiz, if the number of girls is 4 more than the number of boys, then find the number of boys and the number of girls who took part in the quiz.

=> Solution:

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

From first condition,

x + y = 10

From second condition,

y = x + 4

For x + y = 10

x 3 5
y 7 5
(x, y) (3, 7) (5, 5)

 

3 + y = 10

Y = 10-3

Y = 7

5+y = 10

Y = 10-5

Y = 5

For y = x + 4

x 0 2
y 4 6
(x, y) (0, 4) (2, 6)

 

Y = 0 + 4

Y = 4

Y = 2 + 4

Y = 6

∴ The lines intersect at point (3, 7)

∴ The solution of equation is x = 3 and y = 7

∴ The number of boys = 3 and the number of girls = 7

 

 

(iii) 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:

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

From given first condition

5 pencil and 7 pens together cost 50

5x + 7y = 50

From second condition,

7 pencils and 5 pens together cost 46

7x + 5y = 46

For 5x + 7y = 50

x 3 10
y 5 0
(x, y) (3, 5) (10, 0)

 

5(3) + 7y = 50

15 + 7y = 50

7y = 50 – 15

7y = 35

Y = 35/7

Y = 5

5(10) + 7y = 50

50 + 7y = 50

7y = 50 – 50

Y = 0/7

Y = 0

For 7x + 5y = 46

x 3 8
y 5 -2
(x, y) (3, 5) (8, -2)

 

7(3) + 5y = 46

21 + 5y = 46

5y = 46 – 21

5y = 25

Y = 25/5

Y = 5

7(8) + 5y = 46

56 + 5y = 46

5y = 46 – 56

5y = -10

Y = -10/5

Y = -2

∴ The two lines at the point (3, 5)

I.e. the cost of one pencil = 3 and the cost of one pen = 5.

 

 

(iv) Half the perimeter of a rectangular garden is 36m. If length is 4m more than it’s width the find the dimensions of the garden

=> Solution:

Let the breadth of a rectangle be x and the length of the rectangle be y.

Given: Half the perimeter of a garden = 36m

By using formula,

1/2 [2 (l + b)] = 36

L + b = 36

Y + x = 36

X + y = 36

From second condition,

If the length is 4m more than it’s width.

y = x + 4

x – y = -4

For x + y = 36

x 10 20
y 26 16
(x, y) (10, 26) (20, 16)

 

(v) We have a linear equation 2x + 3y – 8 = 0, write another linear equation in two variables x and y such that the graphical representation of the pair so formed is interesting lines. Now, write two linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.

Solution:

Given equation is 2x + 3y – 8 = 0

∴ The lines are interesting

i.e. a1/a2 ≠ b1/b2

∴ Interesting line may be 3x + 7y + 5 = 0

∴ For parallel lines, the condition is,

a1/a2 = b1/b2 ≠ c1/c2

∴ 4x + 6y + 7 = 0 may be parallel line

For coincident lines, the condition is,

a1/a2 = b1/b2 = c1/c2

∴ 8x + 12y – 32 = 0

(vi) The area of a rectangle get reduced by 80sq units, if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units then the area will be increased by sp sq units. Find the length and breadth of the rectangle.

=> Solution:

Let, the length rectangle be x units and the breadth of the rectangle be Y units

Area of the rectangle = l X b = X x Y sq units.

From first condition, by the given problem. The area of a rectangle get reduced by 80 sq units if its length is reduced by 5 units and breadth Is increased by 2 units.

(x – 5) (y + 2) = xy – 80

Xy + 2x – 5y – 10 = xy – 80

2x – 5y – 10 + 80 = 0

2x – 5y + 70 = 0 -+

2x – 5y = – 70

From second condition,

If we increase the length by 10 units and decrease the breadth by 5 units then the area will be increased by 50 sq units.

(x + 10) (y – 5) = xy + 50

Xy – 5x + 10y – 50 = xy + 50

– 5x + 10y = 50 + 50

– 5x + 10y = 100

For 2x – 5y = – 70

x 10 20
y 18 22
(x, y) (10, 18) (20, 22)

 

2 (10) – 5y = -70

20 – 5y = -70

– 5y = -70 – 70

– 5y = – 90

Y = 90/-5

Y = 18

 

2(20) – 5y = -70

40 – 5y = -70

– 5y = -70 – 40

-5y = -110

Y = -110/-5

Y = 22

For -5x + 10y = 100

x 0 20
y 10 20
(x, y) (0, 10) (20 ,20)

 

– 5(0) + 10y = 100

0 + 10y = 100

Y = 100/10

Y = 10

 

-5(20) + 10y = 100

– 100 + 10y = 100

10y = 100 + 100

10y = 200

y = 200/10 = 20

The two lines intersect at the point (40, 30)

∴ The solution is x = 40 and y = 30

∴ Length = 40 units and breadth is = 30 units.

(vii) In a class, if three students sit an each bench, one student will be left, if four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.

=> Solution:

Let, the number of benches be x and the number of students be y.

From first condition,

If three students sit on each bench, one student will be left

y = 3x + 1

From second condition,

If four students sit on each bench, one bench will be left.

y = 4x – 4 The two lines intersect at the point (40, 30)

∴ The solution is x = 40 and y = 30

∴ Length = 40 units and breadth is = 30 units.

 

For 3x – y = -1

x 0 1
y 1 4
(x, y) (0, 1) (1, 4)

 

3(0) – y = – 1

0 – y = – 1

– y = – 1

Y = 1

 

3(1) – y = – 1

3 – y = – 1

– y = – 1 – 3

– y = – 4

Y = 4

For 4x – y = 4

x 2 3
Y 4 8
(x, y) (2, 4) (3, 8)

 

4 (2) – y = 4

8 – y = 4

– y = 4 – 8

– y = – 4

Y = 4

 

4(3) – y = 4

12 – y = 4

– y = 4 – 12

– y = – 8

Y = 8

∴ The lines intersect at point (5, 16)

∴ The solution of the equation x = 5 and y = 16

i.e. the number of benches = 5 and the number of students = 16

 

Here is your solution of Telangana SCERT Class 10 Math Chapter 4 Pair of linear equations in two variables Exercise 4.1

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Updated: June 7, 2021 — 2:31 pm

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  1. Hello
    You have done the 3rd sum wrong once check it out before Sharing with us

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