AP 10th Class Maths 3rd Chapter Pair of Linear Equations in Two Variables Exercise 3.3 Solutions

Well-designed AP Board Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3 offers step-by-step explanations to help students understand problem-solving strategies.

Pair of Linear Equations in Two Variables Class 10 Exercise 3.3 Solutions – 10th Class Maths 3.3 Exercise Solutions

Question 1.
Solve the following pair of linear equations by the elimination method and the substitution method :
i) x + y = 5 and 2x – 3y = 4
ii) 3x + 4y = 10 and 2x – 2y = 2
iii) 3x – 5y – 4 = 0 and 9x = 2y + 7
iv) $$\frac{x}{2}+\frac{2 y}{3}$$ = -1 and x – $$\frac{y}{3}$$ = 3
Solution:
i) Given linear equations are
x + y = 5 …… (1)
and 2x – 3y = 4 ……. (2)
Multiply by;2 of (1) to make the x coefficient are equal.
2x + 2y = 2 × 5
2x + 2y = 10
Subtract (2) from (3)

Therefore, x = $$\frac{19}{5}$$ and y = $$\frac{6}{5}$$

Check :

Put x = $$\frac{19}{5}$$ and y = $$\frac{6}{5}$$ in (2)

ii) Given linear equations are
3x + 4y = 10 ……… (1)
and 2x – 2y = 2 ……. (2)
Multiply by 2 of (2) to make the y coefficients are equal.
4x – 4y = 4 ……. (3)
by adding (1) and (3)

Put x = 2 in (1)
3(2) + 4y = 10
4y = 10 – 6 = 4
y = $$\frac{4}{4}$$
y = 1
Therefore, x = 2 and y = 1

Check :

Put x = 2 and y = 1 in (2)
2(2) – 2(1) = 2
4 – 2 = 2
2 = 2
LHS = RHS
So, our solution is correct.

iii) Given linear equations are
3x – 5y – 4 = 0 and 9x = 2y + 7
3x – 5y = 4 ……(1) 9x – 2y = 7 …….(2)
Multiply by 3 of (1) to make the co-efficients of x equal.
9x – 15y = 12 ……..(3)
Subtract (2) from (3)

Put y = $$\frac{-5}{13}$$ in (1)
3x – 5(-$$\frac{5}{13}$$) = 4
3x + $$\frac{25}{13}$$ = 4
3x = $$\frac{4}{1}$$ – $$\frac{25}{13}$$
= $$\frac{4 \times 13-25}{13}$$
= $$\frac{52-25}{13}$$
3x = $$\frac{27}{13}$$

∴ x = $$\frac{9}{13}$$
Therefore, x = $$\frac{9}{13}$$ and y = $$\frac{-5}{13}$$

Check :

Put x = $$\frac{9}{13}$$ and y = $$\frac{-5}{13}$$  in (2)
9($$\frac{9}{13}$$) – 2($$\frac{-5}{13}$$) = 7
$$\frac{81}{13}+\frac{10}{13}$$ = 7
$$\frac{81+10}{13}$$ = 7

7 = 7
LHS = RHS
So, our solution is correct.

iv) Given linear equations are

Put y = -3 in (1)
3x + 4(-3) = -6
3x – 12 = -6
3x = -6 + 12 = 6

Therefore, x = 2 and y = -3

Check :

Put x = 2 and y = -3 in (2)
3(2) – (-3) = 9
6 + 3 = 9
9 = 9
LHS = RHS
So, our solution is correct.

Question 2.
Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes $$\frac{1}{2}$$ if we only add 1 to the deonominator. What is the fraction ?

ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu ?

iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

iv) Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give her ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received.

v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹ 27 for a book kept for seven days, while Susy paid ₹ 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Solution:
i) Let the numerator x and denominator y.
Then fraction = $$\frac{x}{y}$$
If we add 1 to numerator and subtract 1 from denominator.
Then $$\frac{x+1}{y-1}$$ = 1
x + 1 = y – 1
x – y = -2 ………….(1)
If 1 added to denominator, then
$$\frac{x}{y+1}$$ = $$\frac{1}{2}$$
2x = y + 1
2x – y = 1 ……….(2)
By subtracting (1) from (2)

Put x = 3 in (1)
3 – y = -2
-y = -2 – 3
-y = -5
y = 5
Fraction = $$\frac{x}{y}$$ = $$\frac{3}{5}$$

ii) Let the present ages of Nuri and Sonu are x and y.
Five years ago Nuri age = 3 times of Sonu
x – 5 = 3(y – 5)
x – 5 = 3y – 15
x = 3y – 10 ….. (1)
10 years later Nuri age = 2 times of Sonu
x + 10 = 2(y + 10)
x + 10 = + 20
x – 2y = 20 – 10
x – 2y = 10 …… (2)
Put (1) in (2)
3y – 10 – 2y = 10
3y – 2y = 10 + 10
y = 20
Put y = 20 in (1)
x = 3(20) – 10 = 60 – 10
x = 50
Therefore, Nuri’s present age (x) = 50 years
Sonu’s present age (y) = 20 years

iii) Let the digits in the tens and units place are x and y representively then x + y = 9 ……..(1)
Then two-digited number = 10 × x + 1 × y = 10x + y
If we reversed the digits then number = 10 × y + 1 × x = 10y + x
9 times of number = 2 times of reversed number
9(10x + y) = 2(10y + x)
90x + 9y = 20y + 2x
90x – 2x = 20y – 9y
88x = 11y

8x – y = 0 ……….(2)
By adding (1) and (2)

Put x = 1 in (2)
8(1) – y = 0
8 – y = 0
y = 8
Therefore, the tens digit x = 1 and ones digit (y) = 8
So, number = 10x + y
= 10(1) + 8
= 10 + 8
= 18.

iv) Let the number of ₹ 50 and ₹ 100 notes are x and y respectively.
x + y = 25 ………. (1)
50x + 100y = 2000
50(x + 2y) = 2000

By subtracting (1) from (2)

Put y = 15 in (1)
x + 15 = 25
x = 25 – 15 = 10
Number of ₹ 50 notes (x) = 10
Number of ₹ 100 notes (y) = 15.

v) Let fixed chargees for first three days is ₹ x and additional charges per each day ₹ y.
Sarita paid ₹ 27 per 7 days
= first 3 days + 4 days additional
x + 4y = 27 ……(1)
Susi paid ₹ 1 per 5 days
x + 2y = 21 ……..(2)
By subtracting (2) from (1)

y = $$\frac{6}{2}$$ = 3
Put y = 3 in (1)
x + 4(3) = 27
x = 27 – 12 = 15
Therefore, fixed charges for first three days (x) = ₹ 15
additional charges per each day (y) = ₹ 3.