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

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

(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.

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),

(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

(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
∴ y=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.

(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

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