**KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3** 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 3.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

ex 3.3 class 10 Question 1.

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

(i) x + y = 14

x – y = 14

(ii) s – t = 3

\(\frac{s}{3}+\frac{t}{2}=6\)

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) \(\sqrt{2} x+\sqrt{3} y=0\)

\(\sqrt{3} x-\sqrt{8} y=0\)

(vi) \(\frac{3 x}{2}-\frac{5 y}{2}=-2\)

\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)

Solution:

(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

\(\frac{s}{3}+\frac{t}{2}=6\)

s – t = 3 …………. (i)

\(\frac{s}{3}+\frac{t}{2}=6\) ………… (ii)

From eqn. (i), s – t = 3

-t = 3 – s

t = -3 + s

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

\(\frac{s}{3}+\frac{-3+s}{2}=6\)

\(\frac{2 s-9+3 s}{6}=6\)

5s – 9 = 6 × 6

5s – 9 = 36

5s = 36 + 9

∴ 5s = 45

∴ s=\(\frac{45}{5}\)

∴ 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

\(x=\frac{1.3-0.37}{0.2}\)

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

0.4x + 0.5y = 2.3

\(0.4\left(\frac{1.3-0.37}{0.2}\right)+0.5 y=2.3\)

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

\(y=\frac{0.3}{0.1}=\frac{3}{1}=3\)

Substituting the vlue of ‘y’ in

\(x=\frac{1.3-0.37}{0.2}\)

\(=\frac{1.3-0.3(3)}{0.2}\)

\(=\frac{1.3-0.9}{0.2}\)

\(=\frac{0.4}{0.2}\)

\(=\frac{4}{2}\)

∴ x = 2

∴ x = 2, y = 3

(v) \(\sqrt{2} x+\sqrt{3} y=0\) ………….. (i)

\(\sqrt{3} x-\sqrt{8} y=0\) ………….. (ii)

From eqn (i)

\(\sqrt{2} x+\sqrt{3} y=0\)

\(\sqrt{2} x=-\sqrt{3} y\)

\(\mathbf{x}=-\frac{\sqrt{3}}{\sqrt{2}} \mathbf{y}\)

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

substituting the value of ‘y’ in

\(\mathrm{x}=-\frac{\sqrt{3}}{\sqrt{2}} \mathrm{y}\)

\(=\frac{-\sqrt{3}}{\sqrt{2}} \times 0\)

∴ x = 0

∴ x = 0, y = 0

(vi) \(\frac{3 x}{2}-\frac{5 y}{2}=-2\) ………… (i)

\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\) …………….. (ii)

From eqn (ii)

\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)

\(\frac{x}{3}=\frac{13}{6}-\frac{y}{2}\)

\(\frac{x}{3}=\frac{13-3 y}{6}\)

\(x=\frac{3(13-3 y)}{6}\)

\(x=\frac{13-3 y}{2}\)

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

\(\frac{3 x}{2}-\frac{5 y}{2}=-2\)

\(\frac{3}{2}\left(\frac{13-3 y}{2}\right)-\frac{5 y}{2}=-2\)

\(\frac{39-9 y}{4}-\frac{5 y}{2}=-2\)

\(\frac{39-9 y-10 y}{4}=-2\)

\(\frac{39-19 y}{4}=-2\)

39 – 19y = -2 × 4

39 – 19y = -8

-19y = -8 + 39

-19y = -47

Or 19y = 47

\(y=\frac{47}{19}\)

∴ y = 3

Substituting the value of ‘y’ in

\(x=\frac{13-3 y}{2}\)

\(=\frac{13-3(3)}{2}\)

\(=\frac{13-9}{2}=\frac{4}{2}\)

∴ 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 + c

Solution:

ex 3.3 class 10 Question 2.

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

Solution:

2x + 3y = 11 …………… (i)

2x – 4y = -24 ………….. (ii)

From eqn. (i),

2x = 3y = 11

2x = 11 – 3y

\(x=\frac{11-3 y}{2}\)

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

2x – 4y = -24

\(2\left(\frac{11-3 y}{2}\right)-4 y=-24\)

11 – 3y – 4y = -24 – 11

-7y = – 35

7y = 35

\(y=\frac{35}{7}\)

∴ y = 5.

Substituting the value of ‘y’ in

\(x=\frac{11-3 y}{2}\)

\(=\frac{11-3(5)}{2}\)

\(=\frac{11-15}{2}\)

\(=\frac{-4}{2}\)

∴ x = -2

∴ x = -2, y = 5

Now, y = mx + 3

5 = m(-2) + 3

5 = -2m + 3

5 – 3 = -2m

-2m = 2

\(\mathrm{m}=\frac{2}{-2}\)

∴ m= -1.

ex 3.3 class 10 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.

Solution:

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

\(y=\frac{26}{2}\)

∴ 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.

Solution:

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

\(\mathrm{y}=\frac{162}{2}\)

∴ 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.

Solution:

Let the cost of each bat b Rs. x.

Let the Cost of each ball be Rs. y.

∴ 7x + 6y = 3800

3x + 5y= 1750

From eqn. (ii),

3x + 5y = 1750

3x = 1750 – 5y

\(x=\frac{1750-5 y}{3}\)

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

\(7\left(\frac{1750-5 y}{3}\right)+6 y=3800\)

\(\frac{12250-35 y}{3}+\frac{6 y}{1}=3800\)

\(\frac{12250-35 y+18 y}{3}=3800\)

12250 – 17y = 3800 × 3

12250 – 17y = 11400

-17y = 11400 – 12250

-17y = -850

\(y=\frac{850}{17}\)

∴ y = Rs. 50

Substituting the value of ‘y’ in

\(x=\frac{1750-5 y}{3}\)

\(=\frac{1750-5 \times 50}{3}\)

\(=\frac{1750-250}{3}\)

\(=\frac{1500}{3}\)

∴ 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?

Solution:

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

\(\mathrm{y}=\frac{50}{5}\)

∴ 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 \(\frac{9}{11}\). if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

Solution:

If the Numerator is ‘x’ and denominator is ‘y’, then the Fraction is \(\frac{x}{y}\).

Adding 2 to both Numerator and denominator, then fraction is \(\frac{9}{11}\).

\(\frac{x+2}{y+2}=\frac{9}{11}\)

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 \(\frac{5}{6}\).

\(\frac{x+3}{y+3}=\frac{5}{6}\)

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

\(x=\frac{9 y-4}{11}\)

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

6x – 5y + 3 = 0

\(6\left(\frac{9 y-4}{11}\right)-5 y+3=0\)

\(\frac{54 y-24}{11}-\frac{5 y}{1}+\frac{3}{1}=0\)

\(\frac{54 y-24-55 y+33}{11}=0\)

-y + 9 = 0

-y = -9

∴ y = 9

Substituting the value of ‘y’ in

\(x=\frac{9 y-4}{11}\)

\(=\frac{9 \times 9-4}{11}\)

\(=\frac{81-4}{11}\)

\(=\frac{77}{11}\)

∴ x = 7.

∴ Fraction is \(\frac{x}{y}=\frac{7}{9}\)

(vi) Let the present age of Jacob = x years

and the present age of his son = y years

∴ 5 years hence: Age of Jacob = (x + 5) years

Age of his son = (y + 5) years

According to given condition,

[Age of Jacob] = 3[Age of his son]

x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15

⇒ x – 3y -10 = 0 … (1)

5years ago : Age of Jacob = (x – 5) years,

Age of his son = (y – 5) years

According to given condition,

[Age of Jacob] = 7[Age of his son]

∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35

⇒ x – 7y + 30 = 0 … (2)

From (1) , x = [10 + 3y] … (3)

Substituting this value of x in (2) , we get

(10 + 3y) – 7y + 30 = 0

⇒ -4y = -40 ⇒ y = 10

Now, substituting y = 10 in (3) ,

we get x = 10 + 3(10)

⇒ x = 10 + 30 = 40

Thus, x = 40 and y = 10

⇒ Present age of Jacob = 40 years and present age of his son = 10 years

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 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 3.3, drop a comment below and we will get back to you at the earliest