Pair of Linear Equations in Two Variables | Exercise 3.3

Question 2

  1. 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 12\dfrac{1}{2} if we only add 1 to the denominator. 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.

Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution
Understand the Question

To solve these word problems using the elimination method:

  1. Identify the two unknown quantities and represent them as variables xx and yy.
  2. Translate the two given conditions into a pair of linear equations in standard form (ax+by=cax + by = c).
  3. Multiply equations by constants if needed so that the coefficients of one variable match, then add or subtract the equations to eliminate that variable.
  4. Solve for the remaining variable and substitute it back to find the other variable.

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 12\dfrac{1}{2} if we only add 1 to the denominator. What is the fraction?

Step 1 · Form the Linear Equations

Let the numerator be xx and the denominator be yy. The fraction is xy\dfrac{x}{y}.

Condition 1: Adding 11 to numerator and subtracting 11 from denominator gives 11 x+1y−1=1\dfrac{x+1}{y-1} = 1

x+1=y−1x−y=−2…(1)\begin{aligned} x + 1 &= y - 1 \\ x - y &= -2 \quad \dots (1) \end{aligned}

Condition 2: Adding 11 to denominator gives 12\dfrac{1}{2} xy+1=12\dfrac{x}{y+1} = \dfrac{1}{2}

2x=y+12x−y=1…(2)\begin{aligned} 2x &= y + 1 \\ 2x - y &= 1 \quad \dots (2) \end{aligned}

Step 2 · Solve by Elimination Method

Subtract equation (1)(1) from equation (2)(2) to eliminate yy

(2x−y)−(x−y)=1−(−2)2x−y−x+y=1+2x=3\begin{aligned} (2x - y) - (x - y) &= 1 - (-2) \\ 2x - y - x + y &= 1 + 2 \\ x &= 3 \end{aligned}

Substitute x=3x = 3 into equation (1)(1)

3−y=−2−y=−2−3−y=−5y=5\begin{aligned} 3 - y &= -2 \\ -y &= -2 - 3 \\ -y &= -5 \\ y &= 5 \end{aligned}

Therefore, the fraction is xy=35\dfrac{x}{y} = \dfrac{3}{5}.

Answer

(i) 35\dfrac{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?

Step 1 · Form the Linear Equations

Let Nuri's present age be xx years and Sonu's present age be yy years.

Condition 1: Five years ago, Nuri was thrice as old as Sonu

x−5=3(y−5)x−5=3y−15x−3y=−10…(1)\begin{aligned} x - 5 &= 3(y - 5) \\ x - 5 &= 3y - 15 \\ x - 3y &= -10 \quad \dots (1) \end{aligned}

Condition 2: Ten years later, Nuri will be twice as old as Sonu

x+10=2(y+10)x+10=2y+20x−2y=10…(2)\begin{aligned} x + 10 &= 2(y + 10) \\ x + 10 &= 2y + 20 \\ x - 2y &= 10 \quad \dots (2) \end{aligned}

Step 2 · Solve by Elimination Method

Subtract equation (2)(2) from equation (1)(1) to eliminate xx

(x−3y)−(x−2y)=−10−10x−3y−x+2y=−20−y=−20y=20\begin{aligned} (x - 3y) - (x - 2y) &= -10 - 10 \\ x - 3y - x + 2y &= -20 \\ -y &= -20 \\ y &= 20 \end{aligned}

Substitute y=20y = 20 into equation (1)(1)

x−3(20)=−10x−60=−10x=−10+60x=50\begin{aligned} x - 3(20) &= -10 \\ x - 60 &= -10 \\ x &= -10 + 60 \\ x &= 50 \end{aligned}
Answer

(ii) Nuri is 5050 years old and Sonu is 2020 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 order of the digits. Find the number.

Step 1 · Form the Linear Equations

Let the unit's digit be xx and the ten's digit be yy.

  • Original number =10y+x= 10y + x
  • Reversed number =10x+y= 10x + y

Condition 1: Sum of digits is 99 x+y=9…(1)x + y = 9 \quad \dots (1)

Condition 2: Nine times the original number is twice the reversed number

9(10y+x)=2(10x+y)90y+9x=20x+2y90y−2y=20x−9x88y=11x\begin{aligned} 9(10y + x) &= 2(10x + y) \\ 90y + 9x &= 20x + 2y \\ 90y - 2y &= 20x - 9x \\ 88y &= 11x \end{aligned}

Divide both sides by 1111

8y=xx−8y=0…(2)\begin{aligned} 8y &= x \\ x - 8y &= 0 \quad \dots (2) \end{aligned}

Step 2 · Solve by Elimination Method

Subtract equation (2)(2) from equation (1)(1) to eliminate xx

(x+y)−(x−8y)=9−0x+y−x+8y=99y=9y=1\begin{aligned} (x + y) - (x - 8y) &= 9 - 0 \\ x + y - x + 8y &= 9 \\ 9y &= 9 \\ y &= 1 \end{aligned}

Substitute y=1y = 1 into equation (1)(1)

x+1=9x=9−1x=8\begin{aligned} x + 1 &= 9 \\ x &= 9 - 1 \\ x &= 8 \end{aligned}

The original number is 10y+x=10(1)+8=1810y + x = 10(1) + 8 = 18.

Answer

(iii) 1818

(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.

Step 1 · Form the Linear Equations

Let the number of ₹5050 notes be xx and the number of ₹100100 notes be yy.

Condition 1: Total number of notes is 2525 x+y=25…(1)x + y = 25 \quad \dots (1)

Condition 2: Total amount is ₹20002000 50x+100y=200050x + 100y = 2000

Divide the entire equation by 5050

50x50+100y50=200050x+2y=40…(2)\begin{aligned} \dfrac{50x}{50} + \dfrac{100y}{50} &= \dfrac{2000}{50} \\[0.6em] x + 2y &= 40 \quad \dots (2) \end{aligned}

Step 2 · Solve by Elimination Method

Subtract equation (1)(1) from equation (2)(2) to eliminate xx

(x+2y)−(x+y)=40−25x+2y−x−y=15y=15\begin{aligned} (x + 2y) - (x + y) &= 40 - 25 \\ x + 2y - x - y &= 15 \\ y &= 15 \end{aligned}

Substitute y=15y = 15 into equation (1)(1)

x+15=25x=25−15x=10\begin{aligned} x + 15 &= 25 \\ x &= 25 - 15 \\ x &= 10 \end{aligned}
Answer

(iv) 1010 notes of ₹5050 and 1515 notes of ₹100100

(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.

Step 1 · Form the Linear Equations

Let the fixed charge for the first 33 days be ₹xx and the additional charge per day thereafter be ₹yy.

Saritha's case: Book kept for 77 days (33 fixed days +4+ 4 extra days) x+4y=27…(1)x + 4y = 27 \quad \dots (1)

Susy's case: Book kept for 55 days (33 fixed days +2+ 2 extra days) x+2y=21…(2)x + 2y = 21 \quad \dots (2)

Step 2 · Solve by Elimination Method

Subtract equation (2)(2) from equation (1)(1) to eliminate xx

(x+4y)−(x+2y)=27−21x+4y−x−2y=62y=6y=62y=3\begin{aligned} (x + 4y) - (x + 2y) &= 27 - 21 \\ x + 4y - x - 2y &= 6 \\ 2y &= 6 \\[0.6em] y &= \dfrac{6}{2} \\[0.6em] y &= 3 \end{aligned}

Substitute y=3y = 3 into equation (2)(2)

x+2(3)=21x+6=21x=21−6x=15\begin{aligned} x + 2(3) &= 21 \\ x + 6 &= 21 \\ x &= 21 - 6 \\ x &= 15 \end{aligned}
Answer

(v) Fixed charge is ₹1515 and charge for each extra day is ₹33.

Common Mistakes
  • Two-Digit Representation: Writing a two-digit number as xyxy instead of the place-value expansion 10y+x10y + x.
  • Fixed vs. Total Days in Library Problem: Forgetting that the fixed charge covers the first 33 days, leading to incorrect equations like x+7y=27x + 7y = 27 instead of x+(7−3)y=27x + (7-3)y = 27.
  • Age Problem Offsets: Forgetting to apply the time offset to both persons (e.g., writing x−5=3yx - 5 = 3y instead of x−5=3(y−5)x - 5 = 3(y - 5)).
  • Sign Errors in Subtraction: Making sign errors when subtracting expressions with negative signs, e.g., 1−(−2)=31 - (-2) = 3, not −1-1.

More questions in Exercise 3.3

Q1

Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x+y=5x + y = 5 and 2x−3y=42x - 3y = 4

(ii) 3x+4y=103x + 4y = 10 and 2x−2y=22x - 2y = 2

(iii) 3x−5y−4=03x - 5y - 4 = 0 and 9x=2y+79x = 2y + 7

(iv) x2+2y3=−1\dfrac{x}{2} + \dfrac{2y}{3} = -1 and x−y3=3x - \dfrac{y}{3} = 3

Q2
  1. 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 12\dfrac{1}{2} if we only add 1 to the denominator. 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.

← Back to Pair of Linear Equations in Two Variables