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## Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

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

Answer:

(i) x + y = 5 and 2x – 3y = 4

x + y = 5 → (1)

2x – 3y = 4 → (2)

Multiplying eqn. (1) by 3

3x + 3y = 15 → (3)

By adding eqn. (2) to eqn. (3) ‘y’ is eliminated.

∴ x = \(\frac{19}{5}\)

Substituting the value of ‘x’ in eqn. (1),

x + y = 5

\(\frac{19}{5}+y=5\)

\(y=\frac{5}{1}-\frac{19}{5}\)

= \(\frac{25-19}{5}\)

∴ y = \(\frac{6}{5}\)

∴ x = \(\frac{19}{5}\) y = \(\frac{6}{5}\)

(ii) 3x + 4y = 10 and 2x – 2y = 2

3x + 4y = 10 → (1)

2x – 2y = 2 → (2)

Multiplying eqn. (2) by 2,

4x – 4y = 4 → (3)

Adding eqn. (1) to eqn. (2),

∴ x = \(\frac{14}{7}\) = 2

Substituting the value of ‘x’ in eqn. (1),

3x + 4y = 10

3 × 2 + 4y = 10

6 + 4y = 10

4y = 10 – 6

4y = 4

∴ y = \(\frac{4}{4}\) = 1

∴ x = 2, y = 1.

(iii) 3x – 5y – 4 = 0 and 9x = 2y + 7

3x – 5y – 4 = 0 ⇒ 3x – 5y = 4 → (1)

9x = 2y + 7 ⇒ 9x – 2y = 7 → (2)

Multiplying eqn. (i) by 3,

9x – 15y = 12 → (3)

Subtracting eqn. (3) from eqn. (2)

Substituting the value of ‘y’ in eqn. (1),

3x – 5y = 4

(iv)

3x + 4y = – 6 ……….. (i) 3x – y = 9 → (2)

Subtracting eqn. (2) from eqn. (1),

y = – 3

Substituting the value of ‘y’ in eqn. (1),

3x + 4y = – 6

3x + 4 (-3) = – 6

3x – 12 = – 6

3x = – 6 + 12

3x = 6

∴ x = 2, y = 3

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 if we only add 1 to the denominator. What is the fraction?

Answer:

Let the fraction be \(\frac{x}{y}\), Here numerator is x and denominator is y.

∴ \(\frac{x+1}{y-1}=1\)

x + 1 = y – 1

x – y = – 1 – 1

x – y = – 2 → (1)

\(\frac{x}{y+1}\) = \(\frac{1}{2}\)

2 × x = y + 1

2x = y + 1

2x – y = 1 → (2)

Subtracting eqn. (2) from Eqn. (1),

∴ x = 3

Substituting the value of ‘x’ in Eqn. (1),

x – y = – 2

3 – y = – 2

– y = – 2 – 3

– y = – 5

∴ y = 5

∴ Fraction is \(\frac{x}{y}\) = \(\frac{3}{5}\)

(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 ?

Answer:

Present age of Nuri is ‘x’

Present age of Sonu is ‘y’

Five years ago, age of Nuri is x – 5

Five years ago, age of Sonu is y – 5

then, x – 5 = 3(y – 5)

x – 5 = 3y – 15

x – 3y = – 15 + 5

x – 3y = – 10 → (1)

After 10 years, age of Nuri will be x + 10

After 10 years, age of Sonu will be y + 10

then, x +10 = 2(y + 10)

x + 10 = 2y + 20

x – 2y = 20 – 10

x – 2y = 10 → (2)

Subtracting eqn. (2) in eqn. (1),

∴ y = 20

Substituting the value of ‘y’ in eqn. (2),

x – 2y = 10

x – 2(20) = 10

x – 40 = 10

x = 10 + 40

x = 50

∴ Present age of Nuri, x = 50

Present age of Sonu, y = 20.

(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.

Answer:

Let the two-digit number is 10x + y.

Sum of the digits of to digit number is

x + y = 9 → (1)

Number obtained by reversing the order of the digits, we get 10y + x.

Twicing the 10y + x, it is nine times of first number.

∴ 2(10y + x) = 9(10x + 4)

20y + 2x = 90x + 9y

2x – 90x + 20y – 9y = 0

– 88x + 11y = 0

88x – 11y = 0

8x – y = 0 → (2)

By adding eqn. (i) to eqn. (2),

∴ x = \(\frac{9}{9}\) = 1

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

x + y = 9

1 + y = 9

y = 9 – 1

y = 8

∴ x = 1, y = 8

∴ Two digit number = 10x + y

= 10 × 1 + 8

= 10 + 8

= 18.

(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.

Answer:

Let the ₹ 50 notes be ‘x’

₹ 100 notes b ‘y’

x + y = 25 → (1)

x + 2y = 40 → (2)

Subtracting eqn. (2) from eqn. (1),

∴ y = 15

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

x + y = 25

x + 15 = 25

x = 25 – 15

∴ x = 10

∴ x = 10, y = 15

∴ ₹ 50’s notes are 10 and

₹ 100’s notes are 15.

(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.

Answer:

Fixed charge be ₹ x.

Additional charge be ₹ y.

x + 4y = 27 → (1)

x + 2y = 21 → (2)

Subtracting Eqn. (2) from eqn. (1),

∴ y = \(\frac{6}{2}\) = 3

Substituting the value of ‘y’ in eqn. (1)

x + 4y = 27

x + 4(3) = 27

x + 12 = 27

∴ x = 27 – 12

∴ x = 15

∴ Fixed charge, x = ₹ 15.

Additional charge, y = ₹ 3.