SEBA Class 10 Maths Chapter 3 Exercise 3.3 Solutions – Linear Equations in Two Variables

Q1(i). Solve: \(x + y = 5\) and \(2x - 3y = 4\) by substitution and elimination method.

Answer:
Method 1: Substitution Method
Given:
\[ x + y = 5 \quad ...(1) \] \[ 2x - 3y = 4 \quad ...(2) \]
From (1),
\[ x = 5 - y \quad ...(3) \]
Substituting (3) in (2), we get:
\[ 2(5 - y) - 3y = 4 \] \[ 10 - 2y - 3y = 4 \] \[ 10 - 5y = 4 \Rightarrow 5y = 6 \Rightarrow y = \frac{6}{5} \]
Now substituting the value of y in (3), we get:
\[ x = 5 - \frac{6}{5} = \frac{19}{5} \]

Method 2: Elimination Method

Given:
\(x + y = 5 \quad ...(1)\)
\(2x - 3y = 4 \quad ...(2)\)

Multiply (1) by 2:
\(2x + 2y = 10 \quad ...(3)\)

Now subtract (2) from (3):

     \(2x + 2y = 10\)
     \(-\; (2x - 3y = 4)\)
     -------------------------
     \(5y = 6\)
     \(y = \frac{6}{5}\)

Now substitute in (1):
\(x + \frac{6}{5} = 5\)
\(x = \frac{19}{5}\)

Therefore, \(x = \frac{19}{5},\; y = \frac{6}{5}\)

(ii) \(3x + 4y = 10\) and \(2x - 2y = 2\)

Answer:
Method 1: Substitution Method
Given:
\(3x + 4y = 10 \quad ...(1)\)
\(2x - 2y = 2 \quad ...(2)\)
From (2):
\(2x = 2 + 2y \Rightarrow x = 1 + y \quad ...(3)\)
Substituting (3) in (1):
\(3(1 + y) + 4y = 10\)
\(3 + 3y + 4y = 10\)
\(7y = 7 \Rightarrow y = 1\)
Now from (3):
\(x = 1 + 1 = 2\)

Method 2: Elimination Method
Multiply (2) by 2:
\(4x - 4y = 4 \quad ...(3)\)
Now subtract (1) from (3):

     \(4x - 4y = 4\)
     \(-\; (3x + 4y = 10)\)
     -------------------------
     \(x - 8y = -6\)
Put \(y = 1\):
\(x = 2\)

Therefore, \(x = 2,\; y = 1\)

(iii) \(3x - 5y = 4\) and \(9x = 2y + 7\)

Answer:
Method 1: Substitution Method
From second equation:
\(9x = 2y + 7 \Rightarrow x = \frac{2y + 7}{9} \quad ...(1)\)
Substituting in first equation:
\(3\left(\frac{2y + 7}{9}\right) - 5y = 4\)
\(\frac{2y + 7}{3} - 5y = 4\)
\(2y + 7 - 15y = 12\)
\(-13y = 5 \Rightarrow y = -\frac{5}{13}\)
Now substitute in (1):
\(x = \frac{9}{13}\)

Method 2: Elimination Method
Write second equation as:
\(9x - 2y = 7 \quad ...(2)\)
Multiply first equation by 3:
\(9x - 15y = 12 \quad ...(3)\)
Now subtract (2) from (3):

     \(9x - 15y = 12\)
     \(-\; (9x - 2y = 7)\)
     -------------------------
     \(-13y = 5\)
     \(y = -\frac{5}{13}\)
Substitute in first equation:
\(x = \frac{9}{13}\)

Therefore, \(x = \frac{9}{13},\; y = -\frac{5}{13}\)

(iv) \( \frac{2}{x} + \frac{2}{y} = -1\) and \(x - \frac{y}{3} = 3\)

Answer:
Method 1: Substitution Method
From second equation:
\(x = 3 + \frac{y}{3} \quad ...(1)\)
Substitute in first equation:
\[ \frac{2}{3 + y/3} + \frac{2}{y} = -1 \] Solving, we get:
\(y = -3\)
Now from (1):
\(x = 3 - 1 = 2\)

Method 2: Elimination Method
Given equations are not linear in standard form,
so elimination is not convenient here.
We convert and solve using substitution.

Therefore, \(x = 2,\; y = -3\)

Q2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by 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 numerator be x and denominator be y.
Given:
\[ \frac{x+1}{y-1} = 1 \quad ...(1) \]Also, \[ \frac{x}{y+1} = \frac{1}{2} \quad ...(2) \]
From (1):
\[ x+1 = y-1 \Rightarrow x - y = -2 \quad ...(3) \]
From (2):
\[ 2x = y + 1 \Rightarrow 2x - y = 1 \quad ...(4) \]
Now subtract (3) from (4):

     \(2x - y = 1\)
     \(-\; (x - y = -2)\)
     ---------------------
     \(x = 3\)
Substitute in (3):
\(3 - y = -2 \Rightarrow y = 5\)
Therefore, the fraction is \( \frac{3}{5} \).

(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:
Let present age of Nuri be x years and Sonu be y years.
Given:
\[ x - 5 = 3(y - 5) \quad ...(1) \]Also, \[ x + 10 = 2(y + 10) \quad ...(2) \]
From (1):
\[ x - 5 = 3y - 15 \Rightarrow x - 3y = -10 \quad ...(3) \]
From (2):
\[ x + 10 = 2y + 20 \Rightarrow x - 2y = 10 \quad ...(4) \]
Now subtract (4) from (3):

     \(x - 3y = -10\)
     \(-\; (x - 2y = 10)\)
     ----------------------
     \(-y = -20\)
     \(y = 20\)
Substitute in (4):
\(x - 40 = 10 \Rightarrow x = 50\)
Therefore, Nuri is 50 years old and Sonu is 20 years old.

(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 digits of the number. Find the number.

Answer:
Let the tens digit be x and units digit be y.
Number = \(10x + y\)
Reversed number = \(10y + x\)
Given:
\[ x + y = 9 \quad ...(1) \]Also, \[ 9(10x + y) = 2(10y + x) \quad ...(2) \]
From (2):
\[ 90x + 9y = 20y + 2x \] \[ 88x - 11y = 0 \Rightarrow 8x - y = 0 \quad ...(3) \]
Now subtract (3) from (1):

     \(x + y = 9\)
     \(-\; (8x - y = 0)\)
     --------------------
     \(-7x + 2y = 9\)
From (3):
\(y = 8x\)
Substitute in (1):
\(x + 8x = 9 \Rightarrow x = 1\)
\(y = 8\)
Therefore, the number is 18.

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

Answer:
Let number of Rs 50 notes be x and Rs 100 notes be y.
Given:
\[ x + y = 25 \quad ...(1) \]Also, \[ 50x + 100y = 2000 \quad ...(2) \]
Divide (2) by 50:
\[ x + 2y = 40 \quad ...(3) \]
Now subtract (1) from (3):

     \(x + 2y = 40\)
     \(-\; (x + y = 25)\)
     ------------------
     \(y = 15\)
Substitute in (1):
\(x + 15 = 25 \Rightarrow x = 10\)
Therefore, She got 10 notes of Rs 50 and 15 notes of Rs 100.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for a book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:
Let the fixed charge be x and charge per extra day be y.
For 7 days the extra days will be 4
\[ x + 4y = 27 \quad ...(1) \] For 5 days the extra days will be 2
\[ x + 2y = 21 \quad ...(2) \]
Now subtract (2) from (1):

     \(x + 4y = 27\)
     \(-\; (x + 2y = 21)\)
     ------------------
     \(2y = 6\)
     \(y = 3\)
Substitute in (2):
\(x + 6 = 21 \Rightarrow x = 15\)
Therefore, fixed charge = Rs 15 and charge per extra day = Rs 3.

Q3. The monthly incomes of A and B are in the ratio 9 : 7 and their monthly expenditures are in the ratio 4 : 3. If each saves Rs 1600 per month, find the monthly income of each.

Answer:
Let the incomes of A and B be 9x and 7x.
and expenditures pf Aand B be 4y and 3y.
We know, Savings = Income − Expenditure
Therefore, \[ 9x - 4y = 1600 \quad ...(1) \] \[ 7x - 3y = 1600 \quad ...(2) \]
Now subtract (2) from (1):

     \(9x - 4y = 1600\)
     \(-\; (7x - 3y = 1600)\)
     ----------------------
     \(2x - y = 0 \Rightarrow y = 2x\)
Substitute in (1):
\(9x - 8x = 1600 \Rightarrow x = 1600\)
Income of A = \(9x = 14400\)
Income of B = \(7x = 11200\)
Therefore, incomes are Rs 14400 and Rs 11200.

Q4. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer:
Let number of students in each row be x and number of rows be y.
Total students = xy
\[ (x+3)(y-1) = xy \quad ...(1) \] \[ (x-3)(y+2) = xy \quad ...(2) \]
From (1):
\[ xy - x + 3y - 3 = xy \Rightarrow -x + 3y = 3 \quad ...(3) \]
From (2):
\[ xy + 2x - 3y - 6 = xy \Rightarrow 2x - 3y = 6 \quad ...(4) \]
Now add (3) and (4):

     \(-x + 3y = 3\)
     \(+\; (2x - 3y = 6)\)
     ------------------
     \(x = 9\)
Substitute in (3):
\(-9 + 3y = 3 \Rightarrow y = 4\)
Total students = \(9 \times 4 = 36\)
Therefore, total students = 36.

Q5. A and B each have certain numbers of mangoes. A says to B, “If you give me 30 of your mangoes, I will have twice as many as left with you.” B replies, “If you give me 10, I will have thrice as many as left with you.” How many mangoes does each have?

Answer:
Let A have x number of mangoes and B have y number of mangoes.
\[ x + 30 = 2(y - 30) \quad ...(1) \] \[ y + 10 = 3(x - 10) \quad ...(2) \]
From (1):
\[ x + 30 = 2y - 60 \Rightarrow x - 2y = -90 \quad ...(3) \]
From (2):
\[ y + 10 = 3x - 30 \Rightarrow 3x - y = 40 \quad ...(4) \]
Multiply (3) by 1:
     \(x - 2y = -90\)
     \(3x - y = 40\)
Solve by elimination:
\[ x = 130,\; y = 110 \]
Therefore, A has 130 mangoes and B has 110 mangoes.

Q6. If the pair of equations \(x + ay = b\) and \(ax + y = 1\) has infinitely many solutions, then choose the correct option.

(i) \(a = 1, b = 1\)
(ii) \(a = -1, b = -1\)
(iii) \(a = 1, b = -1\)
(iv) \(a = -1, b = 1\)

(a) Both (i) and (iii)
(b) Both (ii) and (iv)
(c) Both (i) and (ii)
(d) Both (iii) and (iv)

Answer: (c)

Q7. For what value of k, the following pair has infinitely many solutions?

\[ kx + 3y - (k-3) = 0 \] \[ 12x + ky - k = 0 \]

Answer: \(k = 6\)

Q8. Following are the steps of solving a word problem of linear equations in two variables. Choose the correct sequence:

(i) Represent the unknowns using variables.
(ii) Solve the equations.
(iii) Form the required equations.
(iv) Interpret the result.

(a) (i) → (ii) → (iii) → (iv)
(b) (i) → (iii) → (ii) → (iv)
(c) (iv) → (iii) → (ii) → (i)
(d) (ii) → (i) → (iii) → (iv)

Answer: (b)

Q9(i). Solve: \( \frac{2}{x} + \frac{3}{y} = 2\) and \( \frac{4}{x} - \frac{9}{y} = -1\)

Answer:
Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \)
Then equations become:
\[ 2a + 3b = 2 \quad ...(1) \] \[ 4a - 9b = -1 \quad ...(2) \]
Multiply (1) by 2:
\(4a + 6b = 4 \quad ...(3)\)
Now subtract (2) from (3):

     \(4a + 6b = 4\)
     \(-\; (4a - 9b = -1)\)
     ----------------------
     \(15b = 5\)
     \(b = \frac{1}{3}\)
Substitute in (1):
\(2a + 1 = 2 \Rightarrow a = \frac{1}{2}\)
\[ x = 2,\; y = 3 \]
Therefore, \(x = 2,\; y = 3\)

Q9(ii). Solve: \( \frac{1}{2x} + \frac{1}{3y} = 2\) and \( \frac{1}{3x} + \frac{1}{2y} = \frac{13}{6}\)

Answer:
Let \( \frac{1}{x} = a \), \( \frac{1}{y} = b \)
\[ \frac{a}{2} + \frac{b}{3} = 2 \quad ...(1) \] \[ \frac{a}{3} + \frac{b}{2} = \frac{13}{6} \quad ...(2) \]
Multiply (1) by 6:
\(3a + 2b = 12 \quad ...(3)\)
Multiply (2) by 6:
\(2a + 3b = 13 \quad ...(4)\)
Multiply (3) by 3 and (4) by 2:
     \(9a + 6b = 36\)
     \(-\; (4a + 6b = 26)\)
     -------------------
     \(5a = 10 \Rightarrow a = 2\)
Substitute in (3):
\(6 + 2b = 12 \Rightarrow b = 3\)
\[ x = \frac{1}{2},\; y = \frac{1}{3} \]
Therefore, \(x = \frac{1}{2},\; y = \frac{1}{3}\)

Q9(iii). Solve: \( \frac{5}{x-1} + \frac{1}{y-2} = 2\) and \( \frac{6}{x-1} - \frac{3}{y-2} = 1\)

Answer:
Let \( \frac{1}{x-1} = a \), \( \frac{1}{y-2} = b \)
\[ 5a + b = 2 \quad ...(1) \] \[ 6a - 3b = 1 \quad ...(2) \]
Multiply (1) by 3:
\(15a + 3b = 6 \quad ...(3)\)
Now add (2) and (3):

     \(15a + 3b = 6\)
     \(+\; (6a - 3b = 1)\)
     -------------------
     \(21a = 7 \Rightarrow a = \frac{1}{3}\)
Substitute in (1):
\(\frac{5}{3} + b = 2 \Rightarrow b = \frac{1}{3}\)
\[ x = 4,\; y = 5 \]
Therefore, \(x = 4,\; y = 5\)

Q10. In a two-digit number, the sum of the digits is 9. If the digits are reversed, the number is increased by 9. Find the number.

Answer:
Let tens digit be x and units digit be y.
Then number = \(10x + y\)
and number after reverse = \(10y + x\)
\[ x + y = 9 \quad ...(1) \] \[ 10y + x = 10x + y + 9 \quad ...(2) \]
From (2):
\[ 9y - 9x = 9 \Rightarrow y - x = 1 \quad ...(3) \]
Add (1) and (3):
     \(x + y = 9\)
     \(+\; (y - x = 1)\)
     --------------
     \(2y = 10 \Rightarrow y = 5\)
Then \(x = 4\)
Therefore, number = 45.

Q11. The sum of the digits of a two-digit number is 15. The number decreased by 27 is equal to the number obtained by reversing the digits. Find the number.

Answer:
Let digits number be x and y.
\[ x + y = 15 \quad ...(1) \] \[ 10x + y - 27 = 10y + x \quad ...(2) \]
From (2):
\[ 9x - 9y = 27 \Rightarrow x - y = 3 \quad ...(3) \]
Add (1) and (3):
     \(x + y = 15\)
     \(+\; (x - y = 3)\)
     --------------
     \(2x = 18 \Rightarrow x = 9\)
Then \(y = 6\)
Therefore, number = 96.

Q12. The difference between two numbers is 4. Twice the smaller number added to three times the larger number gives 82. Find the two numbers.

Answer:
Let larger number be x and smaller be y.
\[ x - y = 4 \quad ...(1) \] \[ 3x + 2y = 82 \quad ...(2) \]
From (1):
\(x = y + 4 \quad ...(3)\)
Substitute in (2):
\[ 3(y+4) + 2y = 82 \] \[ 3y + 12 + 2y = 82 \] \[ 5y = 70 \Rightarrow y = 14 \]
Then \(x = 18\)
Therefore, the numbers are 18 and 14.