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## Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3

Question 1.

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14

x – y = 14

(ii) s – t = 3

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v)

(vi)

Answer:

(i) x + y = 14 ………. (i)

x – y = 4 …………. (ii)

In eqn. (i), x + 4 = 14

∴ x = 14 – y

Substituting the value of ‘x’ in equation (ii), we have

x – y = 4

(14 – y) – y = 4

14 – y – y = 4

14 – 2y = 4

– 2y = 4 – 14

– 2y = – 10

2y = 10

∴ x = 5

∴ Substituting y = 5 in x = 14 – y,

x = 14 – y

= 14 – 5

x = 9

∴ x = 9, y = 5.

(ii) s – t = 3

s – t = 3 …………. (i)

………… (ii)

From eqn. (i), s – t = 3

-t = 3 – s

t = -3 + s

Substituting the value of ‘t’ in eqn. (ii),

5s – 9 = 6 × 6

5s – 9 = 36

5s = 36 + 9

∴ 5s = 45

∴ s = 9

Substituting the value of s = 9 in t = – 3 + s

t = – 3 + s

= – 3 + 9

t = 6

∴ s = 9, t = 6

(iii) 3x – y = 3 …………. (i)

9x – 3y = 9 ………… (ii)

From eqn. (i), 3x – y = 3

– y = 3 – 3x

y = – 3 + 3x

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

9x – 3y = 9

9x – 3(- 3 + 3x) = 9

9x + 9 – 9x = 9

Here we can give any value for ‘x’ i.e., infinite solutions are there,

y = 3x – 3 Means x = 0 then y = – 3

x = 1 then y = 0

x = 2 then y = 3

etc.

(iv) 0.2x + 0.3y = 1.3 …………. (i)

0.4x + 0.5y = 2.3 …………. (ii)

From eqn. (i)„

0.2x + 0.3y = 1.3

0.2x = 1.3 – 0.3y

Substituting the value of ‘x’ in eqn (ii)

0.4x + 0.5y = 2.3

0.2 (1.3 – 0.3y) + 0.5y = 2.3

2.6 – 0.6y + 0.5y = 2.3

2.6 – 0.1y = 2.3

-0.1y = 2.3 – 2.6

-0.1y = -0.3

0.1y = 0.3

Substituting the value of ‘y’ in

∴ x = 2

∴ x = 2, y = 3

(v) ………….. (i)

………….. (ii)

From eqn (i)

Substituting the value of ‘x’ in eqn (ii)

substituting the value of ‘y’ in

∴ x = 0

∴ x = 0, y = 0

(vi) ………… (i)

…………….. (ii)

From eqn (ii)

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

39 – 19y = – 2 × 4

39 – 19y = – 8

-19y = -8 + 39

-19y = – 47

Or 19y = 47

∴ y = 3

Substituting the value of ‘y’ in

∴ x = 2

∴ x = 2, y = 3

Question 2.

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

Answer:

2x + 3y = 11 …………… (i)

2x – 4y = – 24 ………….. (ii)

From eqn. (i),

2x = 3y = 11

2x = 11 – 3y

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

2x – 4y = – 24

11 – 3y – 4y = – 24 – 11

-7y = – 35

7y = 35

∴ y = 5.

Substituting the value of ‘y’ in

∴ x = – 2

∴ x = – 2, y = 5

Now, y = mx + 3

5 = m(-2) + 3

5 = – 2m + 3

5 – 3 = – 2m

– 2m = 2

∴ m = – 1.

Question 3.

Form the pair of linear equations for the following problems and find their solution by substitution method :

(i) The difference between the two numbers is 26 and one number is three times the other. Find them.

Answer:

Let one number be ‘x’.

another number be ‘y’.

Their difference is 26.

∴ x – y = 26 ………. (i)

One number is three times the other,

∴ x = 3y …………. (ii)

Substituting x = 3y in Eqn. (i),

x – y = 26

3y – y = 26

2y = 26

∴ y = 13

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

x = 3y

∴ x = 3 × 13

∴ x = 39, y = 13

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

Answer:

In two supplementary angles, let one angle be ‘x’.

another angle be ‘y’

Sum of these = 180°

∴ x + y = 180 ………… (i)

The larger of two supplementary angles exceeds the smaller by 18 degrees.

∴ x – y = 18 ………… (ii)

From Eqn. (ii),

x – y = 18

x = 18 + y

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

x + y = 180

18 + y + y = 180

18 + 2y = 180

2y = 180 – 18

2y = 162

∴ y = 81

Substituting the value of y’ in x = 18 + y

x = 18 + 81

x = 99

∴ x = 99°, y = 81°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.

Answer:

Let the cost of each bat b Rs. x.

Cost of each ball be Rs. y.

∴ 7x + 6y = 3800

3x + 5y= 1750

From eqn. (ii),

3x + 5y = 1750

3x = 1750 – 5y

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

12250 – 17y = 3800 × 3

12250 – 17y = 11400

-17y = 11400 – 12250

-17y = -850

∴ y = Rs. 50

Substituting the value of ‘y’ in

∴ x = Rs. 500

∴ x = Rs. 500, Rs. y = 50

∴ Cost of each bat is Rs. 500,

Cost of each ball is Rs. 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

Answer:

Let the fixed charge of the taxi be Rs. x.

If the charge for each km is Rs. y, then

Fixed charge + Charge for 10 km

= Rs. 105

∴ x + 10y = 105 ………. (i)

Fixed charge + Charge for 15 km travelled

= Rs. 155

∴ x + 15y = 155 ……….. (ii)

From eqn. (i),

x + 10 y = 105

x = 105 – 10y

Substituting the value of ’x’ in Eqn. (ii),

x + 15y = 155

105 – 10y + 15y = 155

105 + 5y = 155

5y = 155 – 105

5y = 50

∴ Rs. y = 10.

Substituting the value of y in

x = 105 – 10y,

= 105 – 10 × 10

= 105 – 100

∴ x = Rs. 5

∴ Fixed Charge of Taxi is Rs. 5.

Charge for each km is Rs. 10

Charge for 1 km is Rs. 10

Charge for 25 km ?

∴ 25 × 10 = Rs. 250.

(v) A fraction becomes . if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes . Find the fraction.

Answer:

If the Numerator is ‘x’ and denominator is ‘y’, then the Fraction is .

Adding 2 to both Numerator and denominator, then the fraction is .

11 (x + 2) = 9(y + 2)

11x + 22 = 9y + 18

11x – 9y + 22 – 18 = 0

11x – 96 + 4 = 0……….. (i)

If 3 is added to both Numerator and denominator, then the fraction is .

6(x + 3) = 5 (y + 3)

6x + 18 = 5y + 15

6x – 5y + 18 – 15 = 0

6x – 5y + 3 = 0 ………… (ii)

From eqn. (i),

11x – 9y + 4 = 0

11x = 9y – 4

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

6x – 5y + 3 = 0

– y + 9 = 0

– y = – 9

∴ y = 9

Substituting the value of ‘y’ in

∴ x = 7.

∴ Fraction is

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

Present age of Jacob is ‘x’.

His son’s present age is ‘y’.

After 5 years, the Age of Jacob will be x + 5

After 5 years, the Age of his son will be y + 5

At that time Jacob’s age will be three times that of his son.

x + 5 = 3(y + 5)

x + 5 = 3y + 15

x – 3y = 15 – 5

x – 3y = 10 ………… (i)

5 years ago Jacob’s age was x – 5

5 years ago, the age of Jacob’s son was y – 5

At that time Jacob’s age was seven times that of his son.

∴ x – 5 = 7(y – 5)

x – 5 – 7y – 35

x – 7y = -35 + 5

x – 7y = -30 ………….. (ii)

From Eqn. (i),

x – 3y = 10

x = 10 + 3y

Substituting the value of ‘x’ in Eqn. (ii),

x – 7y = – 30

10 + 3y – 7y = -30

10 – 4y = – 30

– 4y = – 30 – 10

– 4y = – 40

∴ y = 10

Substituting the value of ‘y’ in

x = 10 + 3y

= 10 + 3 × 10

= 10 + 30

∴ x = 40

∴ Present age of Jacob is 40.

Present age of Jacob’s son is 10.