KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4

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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)

$\therefore \quad y=\frac{-5}{13}$
Substituting the value of ‘y’ in eqn. (1),
3x – 5y = 4
$3 x-5\left(\frac{-5}{13}\right)=4$
$3 x+\frac{25}{13}=4 \quad \text { OR } \quad 3 x=4-\frac{25}{13}$
$3 x=\frac{52-25}{13}=\frac{27}{13}$
$3 x=\frac{27}{13}$
$\therefore \quad x=\frac{27}{13} \times \frac{1}{3}$
$\therefore \quad x=\frac{9}{13}$
$\therefore \quad x=\frac{9}{13} \quad y=\frac{-5}{13}$

(iv) $\frac{x}{2}+\frac{2 y}{3}=-1 \text { and } x-\frac{y}{3}=3$
$\frac{x}{2}+\frac{2 y}{3}=-1 \quad \frac{x}{1}-\frac{y}{3}=3$
$\frac{3 x+4 y}{6}=-1 \quad \frac{3 x-y}{3}=3$
3x + 4y = – 6 ……….. (i) 3x – y = 9 → (2)
Subtracting eqn. (2) from eqn. (1),

$y=\frac{-15}{3}=-3$
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
$\therefore \quad x=\frac{6}{3} \quad=2$
∴ 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 $\frac{1}{2}$ 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.