Haryana State Board HBSE 10th Class Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.
Haryana Board 10th Class Maths Solutions 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.
Solution :
(i) By Elimination Method :
The given equations are :
x + y = 5 ………………..(1)
2x – 3y = 4 …………………(2)
Multiplying equation (1) by 3, we get
3x + 3y = 15 ……………..(3)
Adding equation (3) and equation (2), we get
Substituting the value of x in equation (2), we get
2 × \(\frac{19}{5}\) – 3y = 4
\(\frac{38}{5}\) – 3y = 4
– 3y = \(\frac{4}{1}-\frac{38}{5}\)
– 3y = \(\frac{20-38}{5}=-\frac{18}{5}\)
y = \(\frac{-18}{-3 \times 5}=\frac{6}{5}\)
Hence, x = \(\frac{19}{5}\), y = \(\frac{6}{5}\) is the required solution.
By Substitution Method:
The given equations are:
x + y = 5 ……………..(1)
2x – 3y = 4 ……………….(2)
From equation (1), we get
y = 5 – x
Substituting the value of y in equation (2), we get
2x – 3(5 – x) = 4
⇒ 2x – 15 + 3x = 4
⇒ 5x = 4 + 15
⇒ 5x = 19
⇒ x = \(\frac{19}{5}\)
Putting the value of x in equation (1), we get
\(\frac{19}{5}\) + y = 5
y = \(\frac{5}{1}-\frac{19}{5}\)
y = \(\frac{25-19}{5}\)
y = \(\frac{6}{5}\)
Hence, x = \(\frac{19}{5}\), y = \(\frac{6}{5}\) is the required solution.
(ii) By Elimination Method :
The given equations are :
3x + 4y = 10 …………..(1)
2x – 2y = 2
or x – y = 1 …………….(2)
Multiplying the equation (2) by 3, we get
3x – 3y = 3 ……………….(3)
Subtracting equation (3) from equation (1), we get
Substituting the value of y in equation (2), we get
x – 1 = 1
x = 1 + 1 = 2
Hence, x = 2, y = 1 is the required solution.
By Substitution Method: .
The given equations are: –
3x + 4y = 10 …………………(1)
2x – 2y = 2 ………………….(2)
or x – y = 1
From equation (2), we get
x = 1 + y
Substituting the value of x in equation (1), we get
3 (1 + y) + 4y = 10
3 + 3y + 4y = 10
7y = 10 – 3
7y = 7
y = \(\frac{7}{7}\) = 1
Putting the value of y in equation (2), we get
x – 1 = 1
⇒ x = 1 + 1 = 2
Hence, x = 2, y = 1 is the required solution.
(iii) By Elimination Method :
The given equations are :
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 ………………(1)
and 9x = 2y + 7
⇒ 9x – 2y = 7 ……………….(2)
Multiplying equation (1) by 3, we get
9x – 15y = 12 ………………(3)
Subtracting equation (2) from equation (3), we get
By Substitution Method :
The given equations are :
3x – 5y – 4 = 0
⇒ 3x – 5y = 4 ………………(1)
and 9x = 2y + 7
⇒ 9x – 2y = 7 ……………….(2)
From equation (1), we get
3x = 4 + 5y
⇒ x = \(\frac{4+5 y}{3}\)
Substituting the value of x in equation (2), we get
\(\frac{9(4+5 y)}{3}\) – 2y = 7
3 (4 + 5y) – 2y = 7
12 + 15y – 2y = 7
13y = 7 – 12
13y = – 5
y = – \(\frac{5}{13}\)
Putting the value of y in equation (1), we get
3x – 5 × (- \(\frac{5}{13}\)) = 4
3x + \(\frac{25}{13}\) = 4
3x = \(\frac{4}{1}-\frac{25}{13}\)
3x = \(\frac{52-25}{13}\)
3x = \(\frac{27}{13}\)
x = \(\frac{27}{3 \times 13}=\frac{9}{13}\)
Hence, x = \(\frac{9}{13}\), y = – \(\frac{5}{13}\) is the required.
(iv) By Elimination Method:
The given equations are :
Substituting the value of y in equation (2), we get
3x – (- 3) = 9
⇒ 3x + 3 = 9
⇒ 3x = 9 – 3 = 6
⇒ x = \(\frac{6}{3}\) = 2
Hence, x = 2, y = 3 is the required solution.
By Substitution Method :
The given equations are :
\(\frac{x}{2}+\frac{2 y}{3}\) = – 1
\(\frac{3 x+4 y}{6}\) = – 1
3x + 4y = – 6 ………………(1)
and x – \(\frac{y}{3}\) = 3
\(\frac{3 x-y}{3}\) = 3
3x – y = 9 ……………..(2)
From equation (2), we get
3x = 9 + y
x = \(\frac{9+y}{3}\)
Substituting the value of x in equation(1), we get
3 × \(\left(\frac{9+y}{3}\right)\) + 4y = – 6
9 + y + 4y = – 6
5y = – 6 – 9 = – 15
y = – \(\frac{15}{5}\) = – 3
Putting the value of y in equation (2), we get
3x – (-3) = 9
3x + 3 = 9
⇒ 3x = 9 – 3 = 6
x = \(\frac{6}{3}\) = 2
Hence, x = 2, y = – 3 is the required solution.
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?
(ii) Five years ago, Nun was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nun 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 digit. 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, Santa 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 be x and denominator be y, then
The fraction = \(\frac{x}{y}\)
According to question,
\(\frac{x+1}{y-1}\) = 1
⇒ x + 1 = y – 1
⇒ x – y = – 1 – 1
⇒ x – y = – 2 ……………..(1)
and \(\frac{x}{y+1}=\frac{1}{2}\)
⇒ 2x = y + 1
⇒ 2x – y = 1
Subtracting equation (2) from equation (1), we get
Substituting the value of x in equation (1), we
3 – y = – 2
⇒ – y = – 2 – 3 = – 5
y = 5
Hence, the equations are x – y = – 2, 2x – y = 1 and x = 3, y = 5 is the required solution.
Therefore, fraction = \(\frac{3}{5}\).
(ii) Let the present age of Nun be x years and present age of Sonu be y years.
Five years ago age of Nuri = (x – 5) years
Five years ago age of Sonu = (y – 5) years
According to question,
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = – 15 + 5
⇒ x – 3y = – 10 ………………(1)
Ten years later age of Nuri = (x + 10) years
Ten years later age of Sonu= (y + 10) years
According to question,
x + 10 = 2 (y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 20 – 10
⇒ x – 2y = 10 ……………..(2)
Subtracting equation (2) from equation (1), we get
⇒ y = 20
Substituting the value of y in equation (2), we get
x – 2 × 20 = 10
⇒ x – 40 = 10
⇒ x = 10 + 40 = 50
Hence, the equations are :
x – 3y = – 10,
x – 2y = 10 and
x = 50, y = 20 is the required solution.
Therefore,present age of Nuri = 50 years and
present age of Sonu = 20 years.
(iii) Let the digit of unit’s place be x and digit of ten’s place be y
Then required number = 10y + x
After reversing the digits, digit of unit’s place be y and ten’s place be x.
The required number after interchange of digits = 10x + y
According to question, x + y = 9
and 9(10y + x) = 2(10x + y)
⇒ 90y + 9x = 20x + 2y
⇒ 20x – 9x + 2y – 90y = 0
⇒ 11x – 88y = 0
⇒ x – 8y = 0
Subtracting equation (2) from equation (1), we get
Substituting the value of y in equation (1), we get
x + 1 = 9
⇒ x = 9 – 1
⇒ x = 8.
Hence, the equations are x + y = 9, x – 8y = 0 and x = 8, y = 1 is the required solution.
Therefore, the required number = 10 × 1 + 8 = 18.
(iv) Let the number of ₹ 50 notes be x
and number of ₹ 100 notes be y.
Total Amount of ₹ 50 notes = 50x and
total amount of ₹ 100 notes = 100y
According to question,
50x + 100y = 2000
⇒ x + 2y = 40 …………….(1)
and x + y = 25 ……………(2)
Subtracting equation (2) from equation (1), we get
Substituting the value of y in equation (2), we get
x + 15 = 25
⇒ x = 25 – 15
⇒ x = 10
Hence, the equations are x + 2y = 40, x + y = 25 and x = 10, y = 15 is the required solution.
Therefore, ₹ 50 notes = 10
and ₹ 100 notes = 15.
(v) Let the fixed charge be ₹ x
and charge for each extra day be ₹ y
According to question,
x + 4y = 27 ………………(1)
and x + 2y = 21 ……………(2)
Subtracting equation (2) from equation (1), we get
Substituting the value of y in equation (1), we get
x + 4 × 3 = 27
⇒ x + 12 = 27
⇒ x = 27 – 12
⇒ x = 15
Hence, the equations are x + 4y = 27, x + 2y = 21 and x = 15, y = 3 is the required solution.
Therefore, fixed charge = ₹ 15
and charge for each extra day = ₹ 3